How bright is the full Earth during the lunar midnight? At lunar midnight (i.e. the new moon as seen from Earth), the Earth is in its full phase with its entire disk in sunlight, and it is the brightest object in the lunar sky. How bright is it, and how variable is its brightness?

More precisely, I'm interested in specific figures for the magnitude and in comparisons to similar objects. Would I be able to read in that light? To drive? Would I notice the variations in brightness? What are the main variations due to?

Let's take an average albedo for the Earth of 0.3 (it depends, which hemisphere is visible, how much cloud cover etc.). That means the Earth reflects 30% of the light incident upon it.

The flux $f$ falling on the Earth is given by
$$ f_{\odot} = \frac{L_{\odot}}{4\pi d^2} = 1.369\times10^{3}\ Wm^{-2}$$
where $L_{\odot}=3.85\times10^{26}\ W$ from the Sun and $d= 1$AU.

The integrated luminosity from the illuminated hemisphere will be
$$L_{earth} = 0.3\pi R^2 f = 5.2\times10^{16}\ W$$

So now we can compare this with the Sun. One hemisphere of the Sun radiates $1.93\times10^{26}\ W$, and produces a flux of $1.369\times10^{3}\ Wm^{-2}$ at 1 AU. Therefore the illuminated hemisphere of the Earth results in a flux of approximately $f_E=0.056\ Wm^{-2}$, assuming the average Earth-Moon distance of 384,400 km. This calculation assumes isotropic emission, but it is quite likely that the albedo is higher for light reflected through 180 degrees.

The Sun has an apparent magnitude of -26.74, so the magnitude of the "full-earth" at the moon is
$$ m_{Earth} = 2.5\log_{10}\left( \frac{f_{\odot}}{f_{E}}\right) - 26.74 = \underline{-15.77}$$

The answer will of course vary with the albedo of the visible hemisphere, which in turn depends on the time of year and how much of the polar regions can be seen (e.g. http://www.climatedata.info/Forcing/Forcing/albedo.html ). Variations of a few hundredths seem possible, which will lead to apparent magnitude variations in $m_{Earth}$ of $\sim \pm 0.1-0.2$ mag. The albedo may also vary in detail with the exact angle at which the sunlight hits the Earth - an "opposition surge" in brightness, when the Sun-Earth and Moon are almost aligned is possible. The Earth-Moon distance varies from 363,000 to 405,000 km. This will lead to magnitude variations of $\pm 0.12$ mag.

A further way to check this is that the albedo of the Moon is 0.12 and it has a radius of 0.273 times that of the Earth. Therefore the Earth seen from the Moon ought to be $(0.3/0.12)\times(1/0.273)^2 = 33.5$ times brighter. This is 3.81 magnitudes brighter. The mean magnitude of the full Moon is -12.74 (maximum is -12.92), so the brightness of the "full Earth" should be -16.55 on average.

I am not sure why these figures don't agree; I suspect it is that the albedo for reflection when the Sun's light is normally incident on the Moon is quite a bit larger than 0.12. The so-called "opposition surge". If the Earth's albedo behaves in the same way, then the latter figure may be more accurate than my first calculation. My gut instinct is that the answer is somewhere between the two.


Apparent Ring of Craters on the Moon I saw this picture on the BBC news website.

There appears to be a ring of craters in the center of the picture, aligned in an approximate sub vertical orientation.

Am I just seeing things or does this apparent ring of craters a feature and does it extend to the far side of the moon? If it does exist, what form of bombardment or gravitational attraction would have create such a ring?

Actually, most of the surface of the Moon is covered with craters like that. The exception are the large dark 'seas', which are in fact basaltic plains. The seas are mostly present on this side of the moon; the far side of the moon is almost entirely composed of craters.

The reason you see a ring near the terminator (the boundary between the illuminated and dark part) is that the shadows make the craters much more visible.

  Due to the angle at which sunlight strikes this portion of the Moon, shadows cast by craters and other geological features are elongated, thereby making such features more apparent to the observer. This phenomenon is similar to the lengthening of shadows on Earth when the Sun is low in the sky.


Do galaxies have a preferred orientation? This article: http://physicsworld.com/cws/article/news/2011/jul/25/was-the-universe-born-spinning asserts that the universe was "born spinning" and that there is a preferred rotation direction for galaxies in the universe.  My question is whether there is a preferred galaxy orientation in the universe, that is, whether there is a preferred plane for the galaxy disk?

There is no evidence for that. That's either a systematic observation error, or a random result within usual statistical fluctuations, compare this Q&A site.


Spacetime curvature illustration accuracy When a particle travels through spacetime, this is sometimes illustrated moving in flat grid with various depths. How accurate is this way of imagining gravity?

Can the gravity well be infinitely deep?

Gravity well http://blakemaybank.com/wp-content/uploads/2012/12/Gravity_well.gif

Looking at this illustration, it suggests that there is an amount of distance going down the well. Is movement down the well limited by the speed of light? Remember, I'm aware that it is an illustration, and the question is about the accuracy of the illustration.

Is all gravitational wells infinitely deep at the exact center? So that if a particle could be made small enough to fit within the center, it would become a black hole.

Newtonian gravity of a point-source can described by a potential $\Phi = -\mu/r$. If we suppress one spatial dimension and use it to graph the value of this potential instead, we get something that looks very close to this illustration, and is indeed infinitely deep at the center--at least, in the idealization of a point-mass. And farther away from the center, it goes flat, just as many illustrations like this have it.

Illustrations like this are fairly common, and I'm guessing that they're ultimately inspired by the Newtonian potential, because they have almost nothing to do with with the spacetime curvature.

Here's an isometric embedding of the Schwarzschild geometry at an instant of Schwarzschild time, again with one dimension supressed:

Above the horizon (red circle), the surface is a piece of a paraboloid (the Flamm paraboloid). Unlike the potential, it does not go flat at large distances.

Being isometric means that it correctly represents the spatial distances above the horizon at an instant of Schwarzschild time. Below the horizon, the embedding isn't technically accurate because the Schwarzschild radial coordinate does not represent space there, but time. Although if we pretend it's spacelike below the horizon, that would be the correct embedding. Picture the below-horizon part as having one-directional flow into the singularity.

Since we've only represented space and not time, the embedding is not enough to reconstruct the trajectories of particles in this spacetime. Still, it is a more accurate representation of a part of the spacetime curvature of point-source--specifically the spatial part.

  The velocity of the object from this perspective, would seem to increase, until a point - where the velocity in x,y coordinates starts to decrease due to most of the motion happening "down" the time dimension. Is this also correct? Would a photon seem to slow down when moving down the well, if seen from above?

The above is an embedding of a slice of spatial geometry, and is not a gravity well. The mathematical form of the paraboloid above the horizon is best described in cylindrical coordinates as
$$r = 2M + \frac{z^2}{8M}\text{.}$$
Here the vertical $z$ coordinate doesn't mean anything physically. It's purely an artifact of creating a surface of the same intrinsic curvature in Euclidean $3$-space as the $2$-dimensional spatial slice of Schwarzschild geometry.

For the Schwarzschild spacetime, radial freefall is actually exactly Newtonian in the Schwarzschild radial coordinate and proper time, i.e. time experienced by the freefalling object, rather than Schwarzschild time. So the Newtonian gravity well isn't actually a bad picture for the physics--it's just not the geometry and so is not a good representation of how any part of spacetime is curved. For non-radial orbits, the effective potential is somewhat different that than the Newtonian one, but ignoring the effects of angular momentum gets us the Newtonian form.

In Schwarzschild time, yes, a photon (or anything else) does slow down as gets near the horizon. In fact, in Schwarzschild time it never reaches the horizon, which is one indication that the Schwarzschild coordinates are badly behaved at the horizon. The coordinate acceleration actually becomes repulsive close to the horizon--and for a fast enough infalling object, is always repulsive. This can be understood as the particle moving to places with more and more gravitational time dilation. In proper time of any infalling observer, however, close to the horizon the acceleration is always attractive.


Need some sources on radiotelescope sensitivity Could anyone direct me to papers with info on minimal flux that can be registered with radiotelescopes?

I'll try a quick estimate.

The strength of a radio source $S$ is often measured in Jansky's and 1 Jy is $1 \times 10^{-26}$ W/m^2/Hz. That means every square meter of collection area is exposed to $1 \times 10^{-26}$ per Hz of bandwidth.

Total power received would then be

$$P_S = S A \Delta \nu$$

where $A$ is the receiving area of the dish and $\Delta \nu$ is the bandwidth in Hz.

The noise limit of an RF receiving system is often expressed as an equivalent temperature. A typical number might be 50 Kelvin, which includes a receiver temperature, sky temperature, spillover (thermal radio emission radiated from the ground) etc. Though it seems counterintuitive, we can add temperatures as a proxy for adding powers because we are in the Rayleigh-Jeans regime where $\nu << k_B T/h$ as explained in this answer

You convert the noise temperature to a power in Watts:

$$P_N = k_B T \Delta \nu$$

where $k_B$ is the Boltzmann constant which is about $1.381 \times 10^{-23}$ Joules/K, and $\Delta \nu$ is again bandwidth in Hz.

Let's set the minimum strength easily detected to be equal to the noise, and equate the two powers:

$$P_S = P_T $$

$$S A \Delta \nu = k_B T \Delta \nu$$

$$S A = k_B T$$

$$S = k_B T / A$$

Note that the bandwidth drops out since both the thermal noise and most radio sources are fairly broadband. Even if the original line is fairly narrow, you'll tend set your minimum bandwidth of a specific measurement to cover the spectral region of interest.

Ballpark Estimate

The Arecibo dish has a radius of about 150 meters, so the area $A$ is about 70,700 square meters. A typical radio astronomy system temperature is 50 Kelvin (it can be significantly higher or lower depending on many factors, NASA deep space 70 meter dish liquid helium cooled front-end is 20K (see here and here), so plugging that into the equation above, we get $ = 9.8 \times 10^{-27}$ watts, which is (probably not coincidentally) almost exactly 1 Jansky!

Confirm that

According to Joeri van Leeuwen's Radio Astronomy lecture 5, The Radio Telescope at the bottom of page 79, the System equivalent Flux Density or SEFD of Arecibo is 3 Jansky, so a basic back of the envelope calculation gets us quite close!

Other radio telescopes will very roughly scale based on area, a smaller area dish will have a higher SEFD or signal level equivalent to the system noise.

Another example, VLA

A VLA dish is 25 meters in diameter, which is about 150 times less area. My back-of-the-envelope calculation would then be 150 Jansky.

According to Table 3.2.1 on the NRAO page Performance of the VLA during the Next Semester; Sensitivity at its lowest, a VLA dish's best SEFD is 250 Jansky. I've plotted the table below. Still checks out.


What is the closest heavenly body to our solar system What is currently the closest known heavenly body, brown dwarf, rogue planet or other to our solar system? At one time it was Alpha Centauri.

The closest body we currently know to exist is Proxima Centauri which is part of the Alpha Centauri system at about 4.25 light years away.

There may be rogue planets that are closer but we haven't been able to detect any. Objects down to brown dwarf size that are closer than Alpha Centauri have not been found by extensive surveys.

Here is a list of close objects to the solar system: https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs


How does the Sky in Google Earth Pro work? For those who don't know, Google Earth Pro has a feature where you can see what the sky should look like based on the time and geographical location.

I know Google Sky says that Google Earth uses their information, is there any way that I could access this through an API?

If that isn't possible, do y'all know of any related databases that reveal how the sky would look at night? (Emphasis on how the sky would look here, I know there are things like the Yale Bright Star Catalouge, but it doesn't tell me where the milky way would appear in the sky).

Stellarium is an open-source software that has similar features. The code is available on github, and the developer's documentation is here.

Some information about Stellarium scripting can be found in Stellarium simulation and “Real time” Stellarium Scripting?


How can I calculate solar intensity at Saturn I'm trying to work out how big solar panels have to be for a Saturn probe, but can't find any data for solar energy at Saturn.

Saturn is about 10.1 AU from Sun, so by the inverse-square law, its solar illumination is a little less than one-hundredth that of Earth, which is also called the solar constant with a value of 1362 W/m².

$$I = I_0 \frac{1\ \mathrm{AU}^2}{R^2} \approx 1362\ \mathrm{W\ m^{-2}}\frac{1^2}{9.583^2}\approx14.83\ \mathrm{W/m^2}$$

Since Saturn's relatively large eccentricity of 0.0565 puts it anywhere between 9.041 and 10.124 AU from the Sun, $I$ varies from 16.66 and 13.29 W/m^2 during its 29.5 year orbital period.


What is the most oblate astronomical object known? There are some very vast spinning objects out there.   How high can the eccentricity get, or, which I think is easier to understand, how small can the ratio between the polar and equatorial diameter be?

For clarification, I'm asking about objects composed composed of particles (not some boundary or field etc) and that are devoid of non-microscopic empty spaces. I.e. a planet or star etc; not a galaxy or rings or a heliosphere etc.
For the purpose of this question, I just don't find galaxies or nebulae interesting because they can have almost any shape.

I'm talking about structures that are oblate spheroids essentially.

Saturn is the most oblate planet in the solar system. If the equatorial diameter is $a$ and the polar diameter is $b$ then its oblateness, $(a-b)/a = 0.1$.
We do not know the oblateness values for more than one or two exoplanets and even these are somewhat uncertain, but are thought to be lower than Saturn's value. For example http://adsabs.harvard.edu/abs/2010ApJ...709.1219C
The oblateness of a star depends on the ratio of it gravitational acceleration to centrifugal acceleration at the equator. So the most oblate stars ought to be the ones with big radii and fast rotation rates (i.e. types of stars that would have low surface gravities if non-rotating).
Giant stars come in two basic types - red giants, which are evolved intermediate and low-mass stars. These tend to be slowly rotating, though there is a class called FK Com variables which are fast-rotating red giants.
The other class is the blue giants, which are high mass main sequence, or slightly evolved, stars.
For maximum oblateness you want the star with the highest value of:
$$ f = \frac{R \omega^2}{GM/R^2} = \frac{R^3 \omega^2}{GM}$$
where $M$ is the stellar mass and $\omega$ its angular velocity. Often we don;t know $\omega$, but can estimate the rotation velocity at the equator $v=R\omega$. Hence
$$f = \frac{R^3 v^2}{GMR^2} = \frac{Rv^2}{GM}$$
A possible record holder is the O-type star VTFS 102 in the Large Magellanic Cloud.
This probably has a mass of $25M_{\odot}$ a radius of about $10R_{\odot}$ and has a measured rotation speed of $\geq 600$ km/s. Thus its value of $f$ is 0.75, compared with a value of $f=0.15$ for Saturn.
EDIT: Due to Michael B - what about "millisecond" pulsars? The shortest periods are about 1.4 ms, the radii are about 10 km and the masses are likely to be about 1.4$M_{\odot}$. Do the sums and you find $f=0.1$. So, (at least in Newtonian mechanics) they should be no more oblate than Saturn.
A further contender might by the dwarf planet Haumea that orbits at 40-50 au from the Sun. It is thought large enough to have achieved its shape via hydrostatic equilibrium between gravity and internal pressures (and rotation). Its mass is $4\times10^{21}$ kg, (mean) radius is 700 km and it has a rotation period of 3.91 hours. These numbers give $f=0.25$. It has been modelled using this, and matching its light curve, to be a triaxial ellipsoid (rather than oblate) with axis ratios 2 : 1.5 : 1 (http://adsabs.harvard.edu/abs/2006ApJ...639.1238R).


Galactic mass estimation from apparent magnitude and distance Can we estimate the order of the total mass of any galaxy from the apparent magnitude and its distance to us? Just like
\begin{equation}
M_{galaxy}=c\times 10^{8}M_{\odot },
\end{equation}
where $c$ is a constant (can be arbitrary or ignored, the order is important).
For example mass of the Tucana Dwarf of the Local group isn't known. But its apparent magnitude is 15.7 and average distance is $8.8\times10^{5}pc$. How can I estimate its approximate total mass?

First you must estimate the luminosity. That may be difficult unless the magnitude you have is a bolometric magnitude.

Then you assume some mass to light ratio (actually mass to luminosity ratio, where in solar units the M/L of the Sun is 1).

The M/L ratios of galaxies depend on galaxy type (and age, metallicity, star forming history) and vary from about 2-10. However, this is just the stellar mass. In most cases these ratios would need to be increased by a factor of 10 to give the total mass of a galaxy, which may be dominated by dark matter. The total M/L ratio of dwarf spheroidals like Tucana is likely to be $\sim 100$.

In conclusion, there is no single value of $c$ that can convert from a magnitude (luminosity) to a mass. Further reading: https://courses.lumenlearning.com/astronomy/chapter/properties-of-galaxies/


Length of a tidal day or by how much is the Moon late each day? A tidal day is the MEAN time between successive meridian transits of the Moon. (This is sometimes called called a lunar day, a term that should be deprecated for use in this sense, because it is also used for the length of a day on the Moon and should be reserved for the latter meaning to avoid ambiguity.) Its length exceeds a mean solar day of 24 hours by X minutes, which is also the (mean) time-length by which moonrise (or its meridian transit) is later each day than it was yesterday. (Here is an animation by NOAA demonstrating this principle.)

The exact value of X is variously given in different places, ranging from about 48 to 52. However, I cannot find anywhere an authoritative source detailing the "correct" way to calculate X.

Does anyone know of an authoritative method of calculating X, and/or can anyone comment on the three methods given below.

(1) I have a personal theory that puts X at about 48.763 minutes, but I am uneasy about how correct the underlying logic of it is, which is this:

Consider two successive conjunctions (C1 and C2) of the MEAN sun and MEAN moon. At C1, the Sun's azimuth and the Moon's azimuth will, at ALL places on Earth, be the same, and, although this is no longer true for the rest of the lunation, it will once again be true at C2. This is analogous to two clocks: Clock 1 (the MEAN sun) keeps correct time, while clock 2 (the MEAN moon) loses a constant amount of time (X minutes) each day. Since the two clocks agree at C1 and at C2, clock 2 must have lost exactly 24 hours (1440 minutes) during the intervening synodic lunation, whose length (Ln) is 29.5305891203704 days. This suggests that X (the number of minutes that clock 2 loses per day) = (1440 / Ln) = 48.763.

However, I feel uneasy about this logic for this reason: If C1 occurs on April 1 at 06:00:00, UTC, and it is then sunrise at Tema, Ghana (on the prime meridian, near the equator), C2 will be on April 30 at 18:44:03, UTC, and that will not be sunrise at Tema; it will be dusk there, and it will be sunrise at east longitude 169°.

Does this negate the above logic suggesting that X = 24 hours / Ln?

(2) Another method for calculating X suggested to me is this:

It is based on the Moon's sidereal period (Ld), 27.321661 days. The MEAN angular distance along its orbit travelled by the Moon each day is (360° / Ld) = 13.17636° (which equates to a mean orbital speed of about 0.55° per hour.) So, each day, the Earth must rotate 360° plus an extra 13.17636° to point the same meridian back at the Moon again. The Earth's rotational speed is (360° / 24 hrs) = 15° per hour. At that rate, the extra time needed by the Earth to rotate those extra 13.17636° is [(13.17636/360) * 24] = 0.878424 hour = 52.7 minutes.

(3) I argue that even if method (2) is right in principle, it requires two amendments, because:

(a) The Earth rotates 360° in a sidereal day, which is nearly 4 minutes shorter than 24 hours, in which time the Earth rotates nearly 361°, or, more precisely, 360.98565°. (The fraction in the latter value is obtained from the average daily portion of the Earth's orbit: 360°/365.2422 days = 0.985647° per day.) (An animation demonstrating why this is so is here.) So, the extra time needed by the Earth to rotate those extra 13.17636° is:

[(13.17636/360.98565)*24] = 0.8760 hours = 52.56 minutes.

(b) Furthermore, in those extra 52.56 minutes, the Moon has moved on in its orbit by nearly another half a degree. More precisely, it is (360/27.321661)*(52.56/1440) = 0.48094°. So the calculation should be:

[(13.17636+0.48094)/360.98565)*24] = 0.90800 hours = 54.48 minutes.

So we have three different methods of calculating X, yielding values of 48.763, 52.56 or 54.48 minutes.

Comments anyone? Either endorsements or criticisms of any of these three methods will be gratefully accepted.

As you point out, the mean moon conjucts (and actually eclipses)
the mean Sun every M = 29.5305891203704 days. If the moon's sydonic
period were exactly 29 days, the gain would be clearly 1440/29 or
49.6552 minutes. If the moon's sydonic period were exactly 30 days,
the gain would be clearly 1440/30 or 48 minutes. Thus, your
calculation seems to be correct here.

2, 3, 4. I think this calculation is incorrect because you're
confusing sidereal days with solar days, and your attempt to correct
for this is in the wrong direction:

Suppose the moon culminates at noon at a given location (which
technically means the sun also culminates, and we have an eclipse,
as above).
Twenty four hours later, the sun culminates again, and the moon
has moved 13.17636 degrees along the ecliptic in its orbit.
However, the sun has also moved (about 1 degree) along the
ecliptic in the same direction.
In other words, the moon has only "gained" about 12.17636 degrees
with respect to the Sun (it's actually a little more, since the Sun
moves less than one degree along the ecliptic per solar day).
Since we measure time by the sun, we only need to rotate an extra
12.17636 degrees for the moon to culminate.
That's less time than we'd need otherwise, not more.
I'm too lazy, but if you do the calculations keeping the above in
mind, you should get the same answer.


Is any meteorite known to come from a comet? Comets are mostly icy but they should have some rocky components too, right? I wonder if there's any rocky piece of a former comet among collected meteorites. Is it possible today to identify a meteorite as having cometary origin? Or has surviving pieces from comet impacts been too rare (or violent vapourizing) events?

This page -- http://www.amsmeteors.org/fireballs/faqf/ -- suggests that it's very unlikely to find cometary-origin meteorites, even though the majority of observed meteors are cometary in origin, because the latter are too fragile to survive all the way to the ground:

  Based upon photographic fireball studies, cometary meteoroids have
  extremely low densities, about 0.8 grams/cc for class IIIA fireballs,
  and 0.3 grams/cc for class IIIB fireballs. This composition is very
  fragile and vaporizes so readily when entering the atmosphere, that it
  is called “friable” material. These meteoroids have virtually no
  chance of making it to the ground unless an extremely large piece of
  the comet enters the atmosphere, in which case it would very likely
  explode at some point in its flight, due to mechanical and thermal
  stresses.

It goes on to claim that cometary meteoroids make up about 95% of observed meteors, 38% of observed fireballs, and 0% of fresh meteorites. (Since a hypothetical cometary-origin meteorite would weather faster than meteorites of asteroidal origin, they would be even less likely to be found later, and so it's not surprising that the page says 0% of all known meteorites are of cometary origin.)


How many meteors hit the moon every day or in how many days does a new meteor hit the moon? What is the frequency of meteor strikes on the moon per day?

We generally like you to check google before posting questions here.  I posed your question to google and amongst other links, I got these:

http://science.nasa.gov/science-news/science-at-nasa/2006/13jun_lunarsporadic/
http://www.slate.com/blogs/bad_astronomy/2014/02/24/lunar_impact_video_of_an_asteroid_hitting_the_moon.html

Per Bill cook in the top link:

  No one knows exactly how many meteoroids hit the Moon every day. By monitoring the flashes, we can learn how often and how hard the Moon gets hit."

And in the second one:

  The Moon is smaller and has less gravity than Earth, so it gets hit less often than we do.

Both articles mention MIDAS (the Moon Impacts Detection and Analysis System).  Since the first link is from 2006, perhaps soon we will have a good estimate for you.


Does the relative motion between planets other than Venus and Earth form shapes such as the Flower of Venus? I was reading about the Pentagram/Flower of Venus, and how it is caused by a 13:8 ratio between their orbits. I found this GIF particularly fascinating.

My question is, if the ratio changed, would they form other kind of "fractal" shapes? Does this happen between other planets? What would be the condition for the ratio so that the relative motion forms a symmetrical shape like the Cardioid in the case of Venus and Earth?

It boils down to simple remainder mathematics.

If two planets have a 13:8 resonance, then, for every 1 outer planet orbit, the inner planet moves 1 and 5/8ths times around the sun.  Because both planets move in the same direction, that means the inner planet moves ahead 5/8ths of 1 orbit for every outer planet year.

5/8ths in 1 year, means the inner planet passes the outer planet every 8/5ths year, or every 1.6 years.  That's the catch-up time.  The formula is pretty simple:   8/(13-8) or small/(large-small).

That period is known as the synodic period of Venus viewed from Earth.

Throwing out the 13:8, because it's close but not quite right, Venus orbit is 224.701 days and Earth's orbit 365.256 days (sidereal orbits).

224.701/(365.256 - 224.701)
or, 224.701 / 140.555 = 1.59867 Earth years, or 583.92 Earth days.

Venus' synodic period is listed here as 584 days - close enough.

All you need for a repeating pattern is a synodic period that's very close to a divisor of an integer and you always get that when the planets are in resonance.    1.59867 is close enough to 1.6 that 5 Venus synodic years comes very close to 8 earth years.   That's what causes the pattern to repeat.   1.59867 times 5 is very close to 8.   I think it misses by maybe 2.6 degrees every 8 years, but it's still close enough to have gotten the attention and admiration of the ancients.

Venus and Earth aren't really in resonance, just close.  They don't have sufficient tidal forces on each other to maintain a true resonance.

The flower petal you get in your 2nd link, the 5 intersection points or meeting points, those happen in in that Spirograph like image when Earth and Venus are on opposite sides of the sun.  They also happen on a 1.6 earth years and move forward 225 degrees every Earth year.  It looks neat, but like a Spirograph it's a cool design based on nothing more than simple fractions.

As far as fractal, flowers are sometimes used as examples of fractal patterns, so I can see how, looking at the flower pattern, one might think it might be fractal.   It's really not.  Fractals are non-repeating.   Kepler Orbits are repeating.

Some of Earth's Milankovich cycles may be fractal in their variation.   That's another topic though.

@Chappo

using your 7:4

4/(7-4) = 1.333 years to catch up, so that corresponds to 3 passes every 4 years, or, as you said, a triangle.

For resonances and the number of points in the star-pattern, it seems to be a simple matter of subtraction as you said, because the denominator in your example is 7-4, or 3.   Divide by 3, you then need 3 cycles to get back to an integer.


How will planets behave in the night sky as seen by Mars colonists? Certainly among the first colonists on Mars there will be a few people interested in Astronomy enough to enjoy the night sky.

For them, familliar Mars will be missing and there will be a new, unfamiliar planet instead.

Planet by planet, how will their appearance and especially their behavior as seen from Mars differ from the way they behave in the night sky as seen from Earth?

This should be a relatively simple problem to solve - which is the Academician's way of saying "do it as a homework problem" :-) .

First thing: find some celestial mechanics calculator and determine when each outer and inner planet is visible in Mars' night sky.  That gives you timings.

Next thing: compare the distance from each outer planet to Mars vs. Earth. Maybe start with closest and farthest approach as examples. That'll tell you both the angular extent of the planet as well as the relative brightness compared with Terran observations.  Don't forget to add a fudge factor for the atmosphere.

In fact, probably the most obvious difference will be that stars no longer twinkle. (yeah I'm ignoring during serious dust storms).  You could use a wicked large primary and have no need of adaptive optic corrections.

BTW there are photos of Earth as seen by the Mars rovers (or maybe just the surveying satellites, I forget) so you can get an idea of the color/features available.


What would the pressure and temperature of gas be, right above Jupiter's gas/liquid boundary? Older pages like this "Exploration of the Solar System" course page describe the transition as being a few hundred kilometers down.

More recent findings seem to put the boundary deeper. See

  The result was a surprise for the Juno science team because it indicated that the weather layer of Jupiter was more massive, extending much deeper than previously expected. The Jovian weather layer, from its very top to a depth of 1,900 miles (3,000 kilometers), contains about one percent of Jupiter’s mass (about 3 Earth masses).

- nasa.gov, March 7, 2018: "NASA Juno Findings - Jupiter’s Jet-Streams Are Unearthly"

and

  Below whirling jet streams, 3,000 kilometres deep, lies a dense, rotating core of liquid hydrogen and helium.

- abc.net.au, March 7, 2018: "Jupiter: Juno uncovers deep jet streams, dense liquid core — and cyclone mystery"

Given the updated understanding of the mass of gas above it, what would the pressure and temperature be at that level?

Technically there isn't really a gas-liquid boundary because temperatures are well above the critical point of hydrogen (33K and about 18bar). It's a supercritical fluid. There are important changes at various depths though including the bottom of the circulating winds, the transition to metallic hydrogen and (it now appears) an increasing density of dissolved/suspended heavier elements making up a somewhat diffuse core. There is lots of "pre-Juno" information easily accessible, including temperature and pressure at the transition to metallic hydrogen (10000K and 20GPa). This source gives a pressure of 100kBar at the bottom of the circulating winds layer (about 10 GPa), but I can't find a temperature.

I just found this very interesting diagram: from these lecture notes which shows how the expected temperatures and pressures inside all the giant planets relate the various phases of hydrogen. Notice how all of them except Jupiter do cross the short curved black line at the left, so that they do have a gas-liquid boundary at pressures of a few atmospheres and temperatures of a few tens of Kelvin


What should I improve my telescope kit to see more details of Moon? What should I improve my telescope kit to see more details of Moon?
I have the gskyer telescope of 60mm aperture and 350mm focal length.
my eyepieces are 10mm,25mm.
and I have x3 Barlow.
my best viewing of the Moon when I combine the Barlow with 10mm eyepiece.
I want to see deep details of the moon, by the way, this helps me to zoom in more for the planets like Mars, Jupiter, and Saturn.
any help?

The theoretical resolution of a telescope - and the maximum useful magnification that you can use - increases with aperture (That's the diameter of the main mirror or lens, and not the f-ratio (which is what photographers usually mean by aperture)). As a rough guide, the typical everyday maximum magnification is around the same as the aperture in mm (and is achieved with an eyepiece of focal length equal to the scope's f ratio).

Given good seeing conditions (steady atmosphere, which can be rare) you can push the magnification to around twice that (with half the eyepiece focal length, or by using a barlow) - which is what manufacturers usually quote as the maximum magnification.

As magnification goes up, the brightness of the view goes down, since you're spreading the same amount of light over a larger area. (Increasing aperture increases the light gathering area; doubling the aperture catches four times as much light, so increasing the aperture allows brighter views at the same magnification. This is important for fainter objects, but the moon is bright enough that you don't usually have problems even with small scopes)

So the best way to see more detail is to switch to a good quality, larger aperture scope.


Can lightning occur in stars like the Sun? In the Wikipedia article about lightining, the following explanation is given about the electrification process in clouds:

The details of the charging process are still being studied by
scientists, but there is general agreement on some of the basic
concepts of thunderstorm electrification. The main charging area in a
thunderstorm occurs in the central part of the storm where air is
moving upward rapidly (updraft) and temperatures range from −15 to −25
°C (5 to −13 °F); see Figure 1. In that area, the combination of
temperature and rapid upward air movement produces a mixture of
super-cooled cloud droplets (small water droplets below freezing),
small ice crystals, and graupel (soft hail). The updraft carries the
super-cooled cloud droplets and very small ice crystals upward. At the
same time, the graupel, which is considerably larger and denser, tends
to fall or be suspended in the rising air.
The differences in the movement of the precipitation cause collisions
to occur. When the rising ice crystals collide with graupel, the ice
crystals become positively charged and the graupel becomes negatively
charged; see Figure 2. The updraft carries the positively charged ice
crystals upward toward the top of the storm cloud. The larger and
denser graupel is either suspended in the middle of the thunderstorm
cloud or falls toward the lower part of the storm.

This mechanism seems to rule out the possibility of lightning in absence of a condensed phase (liquid droplets/solid crystals). That makes me wonder, is lightning truly impossible in stars hot enough to be fully gaseous, like our Sun? Or could a alternative mechanism generate lightning in such environment.
Even in absence of any alternative mechanisms, I think lightning still could form at least in small, cold stars, or in old brown dwarfs. What would be the size/age threshold?

Stars are not actually gaseous, they are plasmas, i.e., highly ionized. Hence, the entire star is highly conductive and does not easily develop the voltage difference via friction needed for lightning like that in our atmosphere. On the other hand, stellar atmospheres have strong magnetic fields which reconnect, and that causes strong voltage spikes that generate high energy electrons. On the Earth we would call streams of high energy electrons lightning, but on the sun we call them flares.


Radial velocity curves When I measure RV in a program for more spectral lines, will it be equivalent? I mean I measure RVs of a line in blue spectral region, RVs of a line in red spectral region... Many thanks.

It depends what lines you are measuring and in what kind of star.

When you measure the RV from a spectral line, you are measuring an intensity-weighted average RV over the region where the line is formed.

For a star like the Sun, the photospheric lines are all formed within a layer no thicker than about a 1000 km and the differences in RV with depth in the atmosphere are small (though not negligible if you are interested in finding low mass exoplanets using the RVs). On the other hand, if you were to start measuring RVs from lines originating in the chromosphere or Corona then you could find different RVs caused by mass motions of the emitting gas.

Similarly, hot stars are surrounded by fast moving winds with a temperature structure. Different lines can originate at different heights from gas with different velocity. This can lead to different RVs from different lines.

A better way to measure RV is by cross-correlation with a synthetic or observed template of a similar star with a known RV.


What is the physical geometry of this apparent "eclipse" of a tiny moon of Asteroid Florence? Below is a small GIF made from a small subset of images contained in the original 36MB GIF found at https://cneos.jpl.nasa.gov/images/news/florence.p5us.1Hz.s382.sep01.gif as I just found in the Sky and Telescope article Asteroid Florence Has Two Moons.

Although it's tempting to interpret the image as an optical image, I belive this is pulsed-doppler data from the Goldstone Deep Space Network center. I am guessing that one axis is range (delay) and the other is doppler shift.

Still I can't figure out the geometry that would produce an "eclipse" (in this case the moon passing behind the larger asteroid) with it disappearing so "far" from the asteroid.

Could someone help me out here? Is one axis delay (time) and the other doppler shift? One can start with this description but there is more to it than is discussed there.

below: "A radar image shows asteroid 3122 Florence and tiny echoes from its two moons. Here is an animation that shows them more clearly. The direction of the radar illumination (and thus the direction toward Earth) is at the top." From here. NASA / Jet Propulsion Laboratory. This is a small subset of the frames contained in the original 36 MB GIF, and the size has been decreased by a factor of 2 in order to fit in SE's 2 MB limit.

The vertical axis is signal delay, and therefore distance from the radar antenna; this is more or less true spatial information. As the caption indicates, delay is increasing from top to bottom in the image.

The horizontal axis is frequency shift (due to the Doppler effect). To the left are signals with a lower frequency relative to what was sent from the radar antenna, i.e. reflections from points moving away from the antenna; to the right, signals with a higher frequency, coming from points moving towards it.¹ This axis is more difficult to interpret.

For a single rigid body, which Florence presumably is, the only way it can have points moving in different directions at the same time is by rotating. On one edge of the body, points that have just come into view of the radar are moving quickly towards it. This relative movement slows down as the points approach the centre of the radar-visible disk; then, they move away increasingly fast until they disappear from view on the opposite edge. (The speed change isn’t linear, which means the body will be somewhat warped on the radar image.)

Only this effect allows the radar to distinguish between points that are at the same distance. If the asteroid wasn’t rotating at all, or if we happened to look exactly onto one of its poles, it would appear only as a thin vertical line in the image. For a rotating body, note that the radar cannot tell to which direction the rotation is: A mirror asteroid rotating the other way would look exactly the same to it.

Now for the moons. They are not rigidly connected to Florence, and apparently not tidally locked to it either, so they will not follow its rotation. With that, the starting point “moving in different directions at the same time = rotation” which worked on the main asteroid breaks down. When one of the moons is moving towards the radar, that tells us exactly nothing about its relative position to Florence; the visualization is misleading in that respect. Therefore, it is absolutely possible for a moon to be straight behind Florence even when it appears to its side in this particular visualization.

Emily Lakdawalla has a nice blog post on this topic, with animated drawings: How radio telescopes get “images” of asteroids

¹ After correcting for the movement of Florence – or the whole Florence-and-moons system – relative to Earth, of course.


Epsilon Eridani and Sadira I have seen a number of places on the Internet where Epsilon Eridani is referred to as Sadira. (e.g. https://en.wikipedia.org/wiki/List_of_stars_in_Eridanus )   However, it would seem that its an unofficial name and the star got renamed Ran by the IAU in a competition.

How did Epsilon Eridani become known as Sadira by some?
What or who was Sadira?

All star names are unofficial. A few stars have ancient names (such as Sirius) all other stars are referred to by their position in a catalogue or star atlas.

As well as I can tell, Al Sadira means "the ostrich", or perhaps in context "The (riverbank) ostrich" indicating a type of tree that grows by rivers (baby name sites offer "lotus tree").

There may be a list of Arabic star names that lists this star as Al Sadira(h) but the name is not in common use and won't be recognised by most people.

The IAU name, Ran, from that of a Norse river goddess, is a recent innovation, and doesn't seem to be much used. The Bayer Designation "Epsilon Eridani" is by far the most well known.


Can the Moon eclipse Venus? The title basically says it all. As seen from the Earth, is it possible for the Moon to eclipse Venus (or any other planet) or are the orbits inclined such that this never happens?

If such an eclipse is possible, is it a frequent or infrequent event? How would I find out when the next one occurs?

As @Donald.McLean said in comments, the answer is yes, the Moon can and does occult the other planets in the Solar System. When something apparently big (like the Moon) passes in front of something apparently small (another planet) it's called an occultation. (I say apparently because from our perspective the Moon appears larger than the planets.)

The planets lie* in the ecliptic plane whereas the Moon's orbit is inclined at about 5° to the ecliptic plane, so the Moon crosses the ecliptic plane twice each orbit, so it's not as frequent as it would be if the Moon's orbit were in the ecliptic plane.

Occultations of the planets by the Moon can be found here. There is also software (Occult) that you can use to generate your own predictions.

* Although the planets are said to lie in the ecliptic plane, this is a generalisation - they actually have orbits that are within a few degrees of the ecliptic "plane".


Gravitational acceleration inside a planet The gravitational acceleration, g, inside the Earth generally decreases with decreasing distance to the center:

However, apparently for Jupiter, the gravitational acceleration only increases with decreasing distance to its center. Why is this?

You can use Gauss's law for gravitation to work out the gravity as a function of (interior) radius.
$$ \oint \vec{g} \cdot d\vec{A} = -4\pi G \int \rho\ dV\ .$$
What this means is that the flux of gravitational field $\vec{g}$ out of a closed surface is proportional to the mass enclosed within that surface.

Let us assume that the density is a function of (interior) radius, such that $\rho = Ar^{\alpha}$ and that the planet is spherically sysmmetric, so that the left hand side becomes $4 \pi r^2 g(r)$ and the volume element $dV = 4\pi r^2\ dr$. Then
$$ 4 \pi r^2 g(r) = -4\pi G \int_{r=0}^{r} Ar^{\alpha} 4\pi r^2\ dr$$
as long as $\alpha \neq -3$.
$$ r^2 g(r) = - 4\pi G \frac{Ar^{3+\alpha}}{3+\alpha}$$
$$ g(r) = -\frac{4\pi GA }{3+\alpha} r^{1+\alpha}$$
What this shows, is that if $\alpha < -1$, then gravity will increase with decreasing radius.

Now we are in a position to answer your question. In the Earth's deep interior $\alpha \simeq 0$ and the density does not increase greatly as we head towards the centre. This means that $g(r) \propto r$ as is seen in the plot you reference. In the outer part of the crust $\alpha \simeq -1$ and $g(r)$ is roughly constant.

Below is a model density profile for Jupiter (solid line).

Just looking at this I would say that $\alpha$ is just a bit $<-1$ for most of the interior and so the gravity should be increasing slowly as $r$ decreases, but there is a sharp uptick in density at the boundary to the core which would see $\alpha \ll -1$ and the gravity becoming proportional to $r$ to the power of a negative number (i.e. increasing strongly with decreasing $r$), but then must drop to $g=0$ when $r=0$ shortly afterwards.


What's the difference of the oldest light and newest light? Excuse an amateur question, as inspired by this question: "How old is the oldest light visible from Earth?", how to determine the age of light? Does light get old? Does light have characteristic therefore we know this light "A" is different from that light "B" therefore we can track how it going over time? Like tagging animals to observe their activities? How to differentiate one light from the other?

There is no measurable difference between an "old" photon and a "new" one. A photon has a wavelength, a polarisation, and a direction of travel. And no other features.

You may ask if it is meaningful to discuss the age of light. It is certainly possible to say that photons were emitted by a certain body 100 years ago, and these photons are being detected now, and so this light is 100 years old. But it is impossible to measure the age as a property of the light.

We don't know of any characteristic that uniquely specifies the "age" of the light, so we just assume it's travelled from as far away as we think the object that generated it is, and call the time to travel that far it's "age".  In the case of the Cosmic microwave background, this was (we think) released when the universe first became transparent, a few hundred thousand years after the "big bang". The light from this far back has been "stretched" by the expansion of the universe, but not changed by time.

This is normally true of many simple objects. They either exist or not. A proton doesn't age. A neutron may decay, but this is an event, not a process. It is only with complex objects made of many particles that they can change with age while maintaining their identity.


How the position of this black hole known? How do scientists know the positions of black holes?
Observe the highlighted area in the image:

The article is referring to the black hole known as V4641 Sagittarii. It was originally thought to be the closest known black hole to Earth, at 1,600 light-years (0.5 kiloparsecs). It's a binary system which emits X-rays. Original distance measurements were made based on observations during and following an X-ray outburst in 1999. Analysis of the jets emitted and of the inclination of the system led to initial estimates of $0.4$-$1.7\text{ kpc}$ (see Hjellming et al. (2000)), with lower estimates of around $\sim0.5\text{ kpc}$ being favored - the distance that was then commonly quoted.

Later spectroscopic measurements in more detail appear to show these early observations to be incorrect, as Orosz et al. (2000) found. By observing the orbits of the two objects in the system, their masses were determined, which led to an estimate of the companion star's radius. The temperature of the star was determined by examining its spectrum, and its luminosity was computed from this and the radius. By using the color index and taking into account interstellar extinction, the group was able to calculate the distance to the companion star, and thus to the black hole. They found $7.40\leq d\leq12.31\text{ kpc}$, values closer to the currently accepted distance.

An absorption study (Chaty et al. (2001)) found values somewhere in the middle, with $3\leq d\leq8\text{ kpc}$. However, all groups agree that the actual distance to V4641 Sagittarii is much further than 1,600 light-years. The closest known black hole is now thought to be A0620-00, also known as V616 Monocerotis, at around 2,800 light-years (about $0.85\text{ kpc}$) from Earth.

General black hole detection is discussed in the answers to How are black holes found?. The takeaway for this case - a stellar-mass black hole in a binary system - is that effects from the companion star must be observed and measured. Fortunately, there are typically plenty of ways to choose from.


Are we moving faster than the Cosmic microwave background radiation? if the Cosmic microwave background radiation is coming towards us in the speed of light and all matter and energy was singular at the big bang, doesn't that mean we somehow moved faster than the speed of light (otherwise the CMBR would have already passed us not be coming toward us?)

You seem to struggle with a very common thing; the understanding that The Big Bang is not something that happened in space, but rather to space. It could be good for you to read some answers to misconceptions about The Big Bang here, or the answer to a question about it on Physics Stack Exchange. The Big Bang caused everything in the universe to move away from each other; there is no central point from which everything is receding.

For the first few hundred thousand years of our Universe, photons and matter particles were coupled in a photon-particle soup. They were all so close together, that they constantly bumped into each other, not being able to travel far at all before changing directions. As the universe expanded and the temperature of the soup became colder, the photons and matter particles were able to travel further before interacting with each other. When the temperature became low enough for protons and electrons to form hydrogen, the number of matter particles the photons had to interact with became even lower, allowing the photons to travel even further before having to interact with matter particles. It was approximately at this point in time that the CMB originated; the CMB is simply the light of the photons which interacted a very last time around the time photons and matter particles stopped being a photon-particle soup (we call this time the recombination epoch).

The universe was expanding everywhere both at the very start of its existence and during the recombination epoch, as well as now. The release of the CMB photons thus happened everywhere, and the photons of the CMB travelled in all directions. As the universe expands, the photons travel new distances, and we are constantly observing the CMB photons, which simply are reaching us from different places of origin in the universe -- which, for all we know, may be infinitely large, so we may just be seeing these kind of photons until the end of time.


How to see Mars and Venus in color? My telescope is Arsenal - Synta 70/700 AZ2 (refractor model). 140x max zoom (10 mm with Barlow).
I could clearly see Jupiter and Saturn in color. I even saw brown stripes on Jupiter.
Yet, when I tried looking at Venus and Mars they appeared pure white, totally washed out by the Sun. I was observing about hour before sunrise.
Is there a way to see them in true color with my current tech? Maybe they could be seen only at certain dates? Or maybe I need some filters?

You are seeing them in color, it's an issue of the mechanics of your eyes.  Eyes are made of rods and cones, cones seeing color, rods seeing greyscale.  When you are looking through a telescope, your eye adapts to the low light conditions and the rods dominate the sensitivity.
Jupiter appears much larger than both Mars and Venus, so your eye does not adapt as much and the cones relay more information, allowing you to perceive more color for both it and Saturn (which is also larger than both Venus and Mars in our sky).
That said, Mars through a telescope, to me, appears a bright orange.  Venus, to me, is indistinguishable from white, though it should look more of a pale yellow since that is its color in visible light.
You could try looking when the sky is darker, for Mars is visible in the middle of the night.  It's possible that the blue of the sky is affecting your color perception, too.  Venus is always a morning or evening object, so you should try looking at it just a little after it has risen (looking at it just as it rises will result in atmosphere issues).


What does the coordinate of time really mean? The meaning of the $t$ which appears in space-time intervals or metrics in general relativity. I concluded that $t$ was just a mathematical thing which allows us to label the "space-time manifold" and only proper time τ had a physical meaning too. On Wikipedia I also found that the coordinate time is not a time that could be measured by a clock located at the place that nominally defines the reference frame. What does it really mean?

First, imagine that you're in an inertial reference frame. We can think of this as a sort of lattice throughout space, with a synchronized clock at each lattice point in space. Now imagine two events, $A$ and $B$, which occur at space-time coordinates $(t_A, x_A, y_A, z_A)$ and $(t_B, x_B, y_B, z_B)$. The reference frame has (again, synchronized) clocks at $(x_A, y_A, z_A)$ and $(x_B, y_B, z_B)$. The coordinate time $\Delta t_{AB}$ is the difference in time measured by the clock at $B$ and the clock at $A$, so that $\Delta t_{AB}=t_B-t_A$. If $\Delta t_{AB}=0$, the events are simultaneous in that reference frame. However, an observer in a different inertial frame (say, in one with a boost $\vec{\beta}$ in the $x$-direction) might measure a different $\Delta t_{AB}$. This means that coordinate time is frame-dependent - it depends on the reference frame you're in.

Proper time is sometimes called "wristwatch time". Now, there's a single clock, and it measures the time between two events $A$ and $B$ on a single worldline. This proper time $\tau_{AB}$ does not depend on your reference frame, but it does depend on the path you take from $A$ to $B$ - that is, it's path-dependent.

One way to think of the difference is that coordinate time tells you something about one reference frame relative to another (because it's frame-dependent), while proper time tells you something about the worldline of the observer (because it's path dependent).


First time visit to dark sky stargazing site: what do I need? Complete newbie here. I have an opportunity to visit one of the few designated dark sky sites here in Pennsylvania, Cherry Springs State Park. I plan on camping out over night, but I have never gone stargazing before, don't use a telescope, and I don't know what the protocols are as far as binoculars? glasses? Etc. for viewing the night skies. The park allows no lights, save for flashlights covered with red tint. Is there anything else that you more experienced astronomers would suggst that I need to have with me? What should I bring? What should I expect for my first observation tour ever? Thank you.

If you have yet to plan your adventure and the moon is not you main target, be sure to set your date nearest a "New Moon". During this phase the moon will not be visible at night, reducing the light pollution that much more. This will increase the amount of stars you should be able to see. Peak of any meteor shower event would be a bonus aswell.

Red light is also important as you suggested, but keep in mind a cellphone is just as bad as a flashlight. There are free apps available for many devices to reduce your brightness levels even further. Your eyes can take roughly 20 to 30 minutes to adjust to the darkness after exposure to light. This will significantly impact your stargazing experience.

Wether its your first or hundredth time, naked eye observation is always an enjoyable experience. Being that you shall be going to a region with rather dark skies, you should see a suprisingly bright milky way. There should be so many stars in the sky, one could get "lost" looking for often easily  recognized constellations. You may even see meteors, satellites or even the ISS.

That being said, binoculars are not neccesary but are a great tool for the avid or begginer gazer. With a steady hand and a decent pair of binoculars it is possible to see the rings on saturn and moons of jupiter. Many galaxies and starclusters can also be enhanced with the aid of the binoculars. Be sure to lean on something though or it will be way to shaky to see clearly. If you have the space and a half decent pair, I would bring them. If not, no worries the pea soup of stars in the sky should keep you busy.

There are many free resources online that can help you find what your looking for if you have any celestial targets in mind. If you have a mobile device, an astronomy/planetarium app is a very useful tool for any stargazer.

For your first time, the most important thing I can suggest to bring is a comfortable reclining chair or blanket and an open mind. There are so many amazing things to learn about the universe and there is no better place to do it than under the stars.

Also be sure not to forget the bug repellent. Can't see much of the stars if your hiding in your tent. Best of luck and hope you get great weather.


Can we learn anything from observing galaxies edge-on, like the Spindle Galaxy? I was reading about NGC 5866, sometimes referred to as the Spindle Galaxy. It's notable because of its inclination relative to Earth; we see the disk from the side, rather than face-on. The difference is akin to viewing a person's head in profile, instead of looking at their face from the front.

We do lose some information about the morphology of NGC 5866 because of the unusual orientation, as it's difficult to see any spiral arms, bars, or other structures. That said, I'm wondering whether we can still learn more about the galaxy than if we viewed it face-on. For instance, can we gain information about the density distribution of the galaxy at various distances removed from the galactic plane, or perhaps about the structure of its central bulge (if there is one)?

Yes, we can certainly learn things from edge-on galaxies. The easiest things to learn, as you suggest, have to to with the vertical structure of the stars (and gas). For example, our knowledge of how thick galaxy stellar and gas disks are -- and whether this thickness varies with distance from the galaxy center -- comes almost exclusively from studying edge-on galaxies. It is also in many cases easier to study faint stellar and gaseous halos in edge-on galaxies, because in face-on galaxies, these things would be superimposed on -- and largely lost in the glare of -- the face-on disk.

Traditionally, it was also easier to study the radial structure of disks in edge-on systems, because the fact that the line of sight through a disk intersects more stars when the galaxy is edge-on than when it is face-on meant you could make observations at higher signal/noise at large radii. Thus, the detection that many disk are "truncated" (that is, that there is a radius beyond which their brightness falls off more rapidly than it does inside that radius) was first made using photographic data of edge-on spirals.

Some elements of bulge structure have been traditionally easier to study in edge-on galaxies. The realization that some galaxies had "boxy" or "peanut-shaped" bulges -- along with the existence of "cylindrical" off-plane stellar rotation in such bulges and their likely nature as the vertically thickened inner parts of bars -- came about from observations of edge-on galaxies. (Although we now know how to identify such structures in moderately inclined and even face-on galaxies as well.)

(Note that our view of the Milky Way is essentially that of an edge-on galaxy, albeit one seen from the inside.)


How much time passes in the outside universe when falling into a black hole? If someone were to jump into a black hole, looking back at Earth, how far into the future would they see of Earth?  As I understand it, time would dilate approaching infinity as you approach the event horizon.  As I understand, time dilation would reach infinity for an infinitely small amount of time, but Planck time would probably prevent that infinity from arising.  How much time would pass in the outside universe as we fall in?  How far would we see into Earth's future, if we were looking back at it as we fell in?  This is further pondering my prior question and the answers received for:

Would time go by infinitely fast when crossing the event horizon of a black hole?

You are essentially asking the following: if someone falls from the Earth from some way beyond the event horizon of a black hole, how long after they have left can an observer on Earth still signal to them with a light beam?

The answer of course depends on exactly how far the Earth is from the black hole. It is also often forgotten that it is not just light reaching them before the falling observer reaches the event horizon, but there is also a little bit more to add in the (proper) time it takes the falling observer to travel from the event horizon to the singularity.

The answer in all cases is that there is a finite time available, so no, the infalling person does not look back and see the cosmic aeons flying by.

The handwaving reason is that from the perspective of Schwarzschild coordinate time both the falling person and light beams approach the event horizon along an exponential asymptote. The light beams are always travelling faster, so their $dt/dr$ is always shallower and hence it is possible for them to catch the falling observer as long as the delay in sending the light beam does not exceed a finite critical value. Or to put it another way; you can contruct a past light cone from any point on the event horizon (or the space-like singularity) and it encompasses only a finite time anywhere outside the black hole.

The full, very gritty, details were worked out (for a Schwarzschild black hole) at https://physics.stackexchange.com/a/396157/43351

For the case of the person free-falling from a position $r_0$, then the span of time seen by the falling person when looking back towards $r_0$, before they hit the singularity, is given by
$$ c(\Delta t)_{\rm singularity} = r_s \ln \left(\frac{r_s}{r_0-r_s}\right) + \pi r_s\left(\frac{r_0}{r_s} -1\right)^{1/2}\left(1 + \frac{r_0}{2r_s}\right) -r_0,$$
where $r_s$ is the Schwarzschild radius.

Note that as $r_0$ gets bigger then of course $\Delta t$ get bigger and approaches $\Delta t \simeq r_s \pi(r_0/r_s)^{3/2}/2c$, which is just the freefall proper time (i.e. measured on a clock carried by the falling object) for an object to fall into a black hole. i.e. You could see that length of time into Earth's future, but only because you had spent that long falling into the black hole !

Edit: To address Jonathan's question in comments. Take a 10 solar mass BH with $r_s=30$ km. If it takes 1 hour (proper time) to "free fall" radially (formulae above are for radial infall), then $r_0 = 2.42\times 10^{9}$ m. The formula above then shows that $\Delta t$ is 1 hour to three sig figs. The reason for this is that on such a trajectory, the falling object spends a negligible fraction of its infall time anywhere near the black hole, so the GR effects are negligible.


On a log-log plot of surface gravity to planet mass, what is the meaning of the y-intercept? I am playing around with data from exoplanets.org, and am interested in the plot of surface gravity to planet mass. I reproduced this plot  after downloading their data and performed a non-linear regression model to fit the curve. Unfortunately,the covariance matrix of my fit has infinite values, so I am now trying out a linear fit on a log-log plot, shown below.  My fit, for $y=ax+b$, is $a=0.9511$, $b=0.8631$.

I am now thinking about what I am plotting. I suspect there may not be anything interesting in $\log \frac{GM}{R^2}$ vs. $\log M$, but, regardless, I am trying to understand if there is any meaning in the y-intercept.

What are possible explanations for the value $b$?

For planets of constant mean density you have:
$$
M=\rho \times 4\pi r^3
$$
and the surface value of $g$ is:
$$
g(r)=\frac{GM}{r^2}=G \times \rho\times 4 \pi \times r
$$
So for bodies of constant density the surface gravity is proportional to the radius, and the slope as $r \to 0$ tells you the density. So for bodies of equal density $\log(g(r)) \to -\infty$ as $r \to 0$

In terms of mass:
$$
g(M)={G(4\pi \rho)^{2/3}M^{1/3}}
$$
So as $M\to 0$ we have $g(M)\to 0$ and again $\log(g(M))\to -\infty$ and the intercept of the $\log-\log$ plot as $M$ goes to zero alows you to calculate the mean density.


Can circumstances exist such that as seen from some solar system all stars are obscured? I am mostly active on Physics Stack Exchange but a question occurred to me that I think is best submitted here.
This question is about a thought experiment: whether the circumstances of the intended thought experiment are physically possible. For this thought experiment it's not necessary for the circumstances to be physically possible, but it would be nice bonus.

In the history of astronomy the fact that the fixed stars were available as reference of celestial motion was hugely helpful.
For example: my understanding is that Kepler's approach to finding the shape of the orbit of Mars was as follows. From astronomical records extending over a long time the period of the orbit of Mars can be reconstructed. Kepler would take pairs of Mars position observations, spaced one Mars year apart. These pairs of Mars positions were from two different Earth positions, allowing Kepler to triangulate.
The intended thought experiment:
A solar system in a region of its galaxy with such a density of interstellar gas that all the stars are obscured. Without the fixed stars as reference of celestial motion the Kepler problem would be much harder.
(The purpose of this thought experiment is to argue that while it would have been harder to obtain the inverse square law of gravity, it would not be impossible. This can then serve as demonstration as to what theory of motion is.)

Some of my guesses:
My understanding is that all parts of the galaxy can be observed, it's just that for some parts the observed luminosity of the stars is diminished because the light coming from those stars traveled through regions with interstellar gas. That would make it very unlikely that interstellar gas can be dense enough to obscure absolutely all stars.
That brings me to guessing that this thought experiment would require rather contrived circumstances. For instance, maybe it would require a star to be out in intergalactic space. That would mean that star has had the misfortune, in its distant past, of being ejected from its originating galaxy.

[Later edit]
Assuming the planet is rotating and that the axis has an inclination with respect to the orbital plane:
Once the astronomers make the transition to a heliocentric model:
The astromers have the measurement of duration of the solar day, and from the fact that the duration of the year is constant over time they can, it seems to me, infer the ratio of the solar day to the intrinsic day.
The astronomers can opt to use the line of intersection between the plane of the equator and the planet's orbital plane as a reference of zero rotation rate. (Of course, a rotating planet that has an inclination will undergo gyroscopic precession. However, the precession of the equinoxes is slow enough that on the assumption that the reference is stationary highly fertile science will be possible.)
It seems to me that once such a reference system is in place a Kepler has the opportunity to find Kepler's laws of planetary motion.
The purpose of the thought experiment is to demonstrate this: the very motion of the planets of a solar system is sufficient to establish a reference of motion. I submit that the conservation of the angular momentum of a rotating planet is sufficient to provide a solar system wide reference of orientation. Over time astronomers can iterate to an ever more accurate reference of motion.
Summerizing:
Motion itself is sufficient to provide reference of motion.

A planet with no stars?
Our solar system could certainly pass through a Bok Globule, which is a dense cloud of opaque interstellar dust and gas.  They are typically about a light year across, so we could be stuck in one for quite some time.  While this would obscure the other stars while we passed through, I don't think Earth would be a very good place to live during that time.  Gas and dust falling into the Sun would probably strip or change our atmosphere and end life on Earth.
When stars form in a Bok Globule, the gas in the star's sphere of influence gets pulled into the interstellar disk and then either absorbed by the star and planets or ejected out of the stellar system.  I suspect that a star couldn't form without leaving big holes in the globule.
But suppose life emerged on a tidally locked planet. If only the sun-facing side of the planet was warm enough to support life, but the dark facing side was fatally cold (even near the day/night transition zone), then the inhabitants might never see the other stars due to light pollution from their sun.
Another possibility could be that a planet was in a binary star system.  If the primary star was much more massive than the other star, and the planet was in an L1 orbit with the less massive star around the more massive star, then even if the planet was turning, there would be no night.  Here is a picture from wikipedia to illustrate the orbit.

Unfortunately the L1 orbit is unstable, so a planet in such an orbit wouldn't stay there for long.
On a planet in constant daylight, the other planets would also be difficult or impossible to see.  Highly reflective gas giants in nearby orbits would certainly be visible, though.
Orbital calculations without a fixed celestial sphere?
It would be much harder to deduce a model for the motion of the planets without fixed stars in the back ground.
But say one has a single observatory, and the only things one measures are the times and azimuth angles for the high magnitude solar system bodies (the Sun, Moon, Mercury, Venus, Mars, Jupiter, and Saturn) as they ascend into the sky and descend out of it (for a fixed elevation angle).
If one gathers these for a few years, they will have enough data to have an overdetermined system to calculate the latitude of my observatory, the Earth's axial tilt, the Earth's rotation rate, and the orbital elements of all the bodies they can observe.
One could do it as Kepler did, by matching geometric shapes to the observations.  Better yet, one could do it as Newton did, with a theory of gravity.


What makes the Earth infinitely rotating around its axis I was reading two sites before I posted the question here. One site says something like that the collisions in Universe eventually set the planets rotating, another site says that our planet is formed from some collapsing Nebula or whatsoever and Nebula was rotating when it was about to die and that set the Earth rotating at first place.

The question should be probably better specified. Is Earth rotating because of some continuous force, or because at first place there was some one-time force which set it rotating and the effect lasts until now?

As for the first case, which force is it? As for the second case, that should break our physical laws, because there would be one force, applied a very long time ago which is still in effect and not about to slow down and with a very very long lasting effect. Because thousands or millions of years for rotation is a very long time, isn't it. Or is it slowing down?

BTW: the infinite movement around orbit is bit mysterious too, because you can say that's because of gravity, but gravity itself attracts the things only to one direction, but orbit movement is like parallel with the gravity emitter, and again it is perpetual.

As for the second case, that should break our physical laws

That is precisely the wrong statement. The fact that the Earth is still spinning matches our physical laws exactly. The physical law I'm specifically referring to is known as Newton's First Law of Motion, which is loosely paraphrased as:

  An object at rest remains at rest unless acted upon by a net, outside force. Conversely, an object in motion remains in motion unless acted upon by a net, outside force.

What this means is that once something starts moving (like the Earth spinning) it will keep moving until something else stops it. The Earth long ago started spinning because of the initial angular momentum of the collapsing cloud which eventually made up the Earth. There is nothing in space to stop the Earth from spinning so it is just going to go on spinning1.

You appear to believe that if you start an object moving by applying some force, it will eventually run down and stop moving of its own accord. This is a common thought, based on everyday experiences. In fact, that was what Aristotle thought (and much of the world up until Isaac Newton's time). But what you're not taking into account is that there's always a force being applied to slow things down. Consider a car rolling along the highway. If you take your foot off the gas pedal, the car starts slowing down. But it isn't because some physical law says it should, it's because a net force is acting on the car to slow it down $-$ that net force being wind resistance on the car. So you see, even common, everyday experiences hold up to Newton's first law.

1Technically there are some small forces acting on the Earth to slow it down but those are nearly negligible and mostly from the Moon. These forces would take billions of years or longer to make the Earth come to a complete stop (if they ever even could).


Do heavier elements breakdown during supernova? Heavier elements like gold, uranium etc. are formed at the end of a star's life. As the star explodes into a supernova, it gives rise to nebula which is the birthplace of new stars. But as the star has already fused lighter elements into heavier elements, where does the new hydrogen required for the formation of new stars come from? Do heavier elements breakdown during supernova or is it some other process?

The answer is that in a pre-supernova star, most of its mass is still in the form of hydrogen and helium. It is only the central core where the primordial H and He has fused to heavier elements.

This picture of onion layers is typically what you see in elementary text books. It is completely misleading in a quantitative sense. It schematically represents what's at the centre of the pre-supernova star, but in terms of the mass that is in each shell (it is obviously a 3d object) you get completely the wrong idea, because this diagram is only about 1 Earth diameter across, compared with the actual stellar radius of something like the distance between the Earth and the Sun!

Here is a more sophisticated plot taken from a paper by Fuller et al. (2015). It shows time until the supernova along the x-axis and the y-axis represents a radial assay of the chemical composition from the centre of the star to the outside. The initial total stellar mass is $12M_{\odot}$. As you move leftwards towards the supernova explosion, notice how what is at the core changes - from being H dominated, to He dominated, to C/O dominated then Si and finally Fe (actually iron-peak elements). Note how much mass is contained within these core region for each stage of nuclear burning. The edge of the "helium core" encloses the central $4M_{\odot}$ of the star. The subsequent heavier element cores inside the onion ring structure enclose significantly less mass, until the iron core is around $1.3M_{\odot}$ just prior to the explosion. Blue shading indicates regions that are thoroughly mixed and homogenised by convection.

After the explosion, the neutron star that is produced will also have a mass of around $1.3-1.4M_{\odot}$. In other words most of the rest of the star (about $10M_{\odot}$ just prior to the explosion) gets blown out in the supernova. But of the $8.6M_{\odot}$ that makes into the interstellar medium, well over half is still in the form of hydrogen and helium; the minority will be carbon, oxygen, neon, silicon, iron etc., and only a very small fraction of that will have been transformed (by the r-process) into elements heavier than iron and nickel.

Thus although the material injected back into the interstellar medium is enriched with heavier elements, there is still plenty of hydrogen to start a new generation of stars. It is also the case that star formation is an inefficient process, so the material from which the supernova progenitor formed will still mostly be around in the interstellar medium. The picture you should have is of a gradual enrichment with heavy elements, especially as the interstellar medium gets churned up and mixed through a variety of processes (including supernova explosions!).

EDIT: Here is an even more awesome picture. The lower plot shows the relative mass fraction of each element as a function of enclosed mass as you work your way out from a 15 solar mass star (it has shed 2 solar masses during its evolution). The really awesome thing is that it is animated, so it shows you the first few moments after the core-collapse and how things start to change. Note that the outer $5M_{\odot}$ of the envelope is about half H and half He by mass prior to the core collapse. Lots of He and then O in the layers below that. The upper plot show how the density temperature and outward velocity are behaving. The image is from the website of Woosley and Heger (2007), a canonical work on the subject.

Hmm. I can't upload the animated gif. Here it is; well worth a look.


If the Higgs field only formed after the Big Bang, how was hydrogen formed? These two statements seem to conflict, but since we're here they can't:

Electrically charged particles cannot be massless.
Hydrogen was created at the moment of the Big Bang.

However, the Higgs field theory states that the field was only created AFTER the Big Bang. This implies that all particles from which hydrogen could be formed were massless. Hydrogen consists of 2 charged particles.

If you could provide an answer without complex math I would really appreciate it.

Hydrogen was not "created at the moment of the big bang".

Particles (leptons and quarks) can attain mass via the Higgs field after the epoch of electroweak symmetry breaking, that occurred about a picosecond $(10^{-12}$s) after the big bang. Only after this was it possible to form the building blocks of hydrogen.

In fact the formation of stable protons comes "much" later at about a millionth of a second. Most of the mass of a proton does not come about from the Higgs field, but from the binding energy of its constituent quarks.

Hydrogen (the atom, rather than the proton nucleus) forms much, much later; about 400,000 years after the big bang.


Pointing direction movement of a fixed camera first doubt here:

I have a fixed camera, whose coordinates I know, pointing to the sky. After the first image, I performed a astrometry calculation from nova.astrometry.net and I got, among other things, the Ra and Dec of the center of view. I would like to use the calibration information for other pictures without needing to calibrate again, e. g. copy the calibration info and just update Ra and Dec of the next pictures.

I thought I could do so transforming the Ra and Dec to Az and Alt using the Julian Date of the first event(the calibrated one), and then go back to Ra and Dec using the Julian Date of the second event. But I get wrong results. I tried between different pictures and sometimes the error is bigger some time smaller.

I am using the functions eq2hor and hor2eq of Astrolib (IDL), which takes in account precession and nutation, so I assume than there is a gap in my knowledge and procedure, rather than a programming mistake.

What am I doing wrong?

Thanks in advance

As has been noted, the camera, if truly fixed, will point at spot with fixed declination. The RA it points to will be a simple function (namely so many hours and minutes ahead/behind) of the local sidereal time.

Either your camera is not truly fixed (a bit of shake) or (possibly more likely), you have not correctly determined the local sidereal time.


When did the concept of "celestial sphere" fall? I'm just done reading The Great Ptolemaic Smackdown, and it provided a good timeline of establishing heliocentrism, broad acceptation of Kepler's system, and the initial (failed) Galileo's experiment attempting measurement of parallax of distant stars (which was based on assumption of two near stars of different brightness being a long distance apart; Galileo unfortunately picked a binary star) - all the models described within assume a fixed, mostly immutable "celestial sphere" of distant stars.

When, and how was this concept disproven?

Friedrich Bessel's 1838 observation of stellar parallax disproved the notion that all stars were equally distant.

However, by this time, the idea that the stars were on a sphere surrounding the solar system had already been discounted (but not disproven) by the application of the Copernican principle (The Earth does not have a privileged position) and Occam's Razor (Adopt the model that has the fewest assumptions)

Giordano Bruno is particularly noted with the idea (c. 1580) that the sun is just one star among many which are distributed throughout the universe.


How to deal with the celestial meridian? The book “Foundations of Astrophysics” by Ryden and Peterson states that one measures azimuth in the horizontal coordinate system eastwards starting on the northernmost point of the horizon circle. Thus, east becomes 90 degrees, et cetera. Wikipedia and other sites back this up.

However, it seems as if the position of the vernal equinox is always given relative to the southern part of an observer’s celestial meridian. During the vernal equinox, the sun culminates in the south, setting the siderial time to 0h0min. Knowing the difference between the local siderial time and a star’s right ascension gives you the current position of a star, relative to the southern half of the celestial meridian.

This website describes this too: http://www.polaris.iastate.edu/NorthStar/Unit4/unit4_sub2.htm

Did I miss something? We so explicitly defined azimuth as being relative to the most northern point on the meridian. Now we’re doing all our measurments relative to the south. I’m not complaining: stars culminate in the south, so this makes it easier. But where did we switch? And when must I switch? Did I make a mistake somewhere?

I think you are overlooking these factors:

Iowa State University is at 42 degrees North latitude, so the Sun and celestial equator that they are showing in their figures are always to the south at the meridian. Naturally, observers in the southern hemisphere would see the sun and celestial equator to their north at the meridian.
The lines of right ascension go from pole-to-pole.
So the hour angle that the website is describing is the angle between the plane of the meridian and the plane of right ascension. This angle applies regardless of where you look in the sky: north, south, or overhead.


How can a singularity actually exist? Probably a stupid question, but it "bothers" me and I can't find a concrete answer on the internets.

When you fall into a black hole, afaik, time slows down for the matter falling in as it gets blasted by high energy light incoming as the universe speeds up, relative to that matter.

Doesn't that mean though that the matter falling in never completely reaches the center?

It seems that the "core" of the black hole would be made of whatever was there at the time of the collapse with time not really flowing for it, and incoming matter would slow down as it fall and be in a almost suspended state as it gets closer to center through the event horizon.

But again, all my knowledge here is based on random YouTube, so I'd love to learn more about it if there are any resources touching that aspect...

Thanks!

The answer is, of course, that a singularity can't exist.  The fact that General Relativity predicts that singularities are inevitable under certain conditions is proof that GR is wrong. A singularity is an infinity and a place where the equations of GR literally break down and become meaningless. (Infinities are decidedly unphysical!) When (if!) a more complete theory of space, time and gravity is developed, the new theory will probably show finite results where GR shows a singularity forming.  (I say probably because whatever more complete theory of gravity we develop next is not guaranteed to be the final, completely correct theory -- it may just be an improvement which solves some issues with GR, but not all.)

There is good reason to think that the successor to GR will reproduce nearly all of GR's predictions -- GR is, after all, a very well verified theory.  No prediction of GR that we have been able to test has been less than 100% right, and any new theory of gravity will have to match GR in those realms while showing differences in places (like the center of Black Holes) where gravity is very, very, very..., very strong.

Now the existence and properties of a BH's event horizon is a (no pun intended) gray area.  Unlike near the singularity at the center of a GR black hole, there is no obvious signal in the mathematics of GR that it fails there.  Nor have we made any measurements of the properties of an EH to test against GR's predictions. So our confidence in those predictions is due to the believed-high quality of the theory rather than due to test successfully passed.

(In fact, there's good reason to think that any post-GR theory of gravity will make different predictions about the nature of the event horizon, since it's very difficult to  reconcile the GR's event horizons with quantum mechanics, and quantum mechanics will almost certainly will be part of the next theory of Gravity.)

Having said all that, there's a second answer to your question, that the question itself is ill-formed. As @mmeent says in a comment to another answer, you are assuming a flat space-time and stationary observers, needed in order that simultaneity between those distant observers and observers near the BH be meaningful:  It is meaningful to talk about what any observer sees, but not meaningful to speak about what a distant observer and an observer near the event horizon see at the same time.

There is no inconsistency with observers at infinity seeing things slow down asymptotically as they approach the EH so as to never quite reach it, and observers falling into the same BH passing the EH in a finite time (and noticing nothing special about space at the EH).  The whole point of the distant observer seeing nothing ever quite reach the EH is a consequence of there being no timelike or null curves (light follows null curves and matter always follows timelike curves) which reach out of the BH to outside observers.  You can't get here from there.

It's also worth remembering that nearly every statement you have ever read about BHs is about equilibrium BHs. But a BH where the mass is falling in asymmetrically isn't going to be at equilibrium. Fortunately, GR is a dynamic theory and GR's BHs can be dynamic objects, though they settle down to equilibrium very, very quickly.  We have observed this process using LIGO (the ringdown of two merging BHs is almost the poster child for "Not At Equilibrium") and the observations appear to be consistant with the predictions of GR. (This is a hot area of research, so new, surprising, results are very possible.)

I'd also note that if you fall into BH, the incoming blue-shifted light won't fry you because you, yourself will be falling along with the light.  What an observer at infinity sees as a huge flux of high-energy radiation as it falls close to the EH, you, falling with it, will see as ordinary starlight. (If you could somehow stand still, you'd see the blue-shifting, though.)  If you like, you can think of your high-speed fall red-shifting that blue-shifted light back to normal.


Do Mercury and Venus exhibit retrograde motion? This is more a terminology question, since Mercury and Venus certainly "move backwards" in the sky. The question is whether we call this retrograde motion. I'm referring to apparent retrograde motion here, not retrograde rotation.

I'm under the impression we don't, and reserve that name only for planets that are further away from the Sun than us; however conclusive evidence for whether this is (or isn't) the widely-accepted meaning for "retrograde motion" is hard to find.

As noted by PM2Ring, the notion of "retrograde motion" predates the "Copernican revolution" So you should think about the motions of the planets from a Ptolemaic view:

  The planets have two major motions (in addition to the diurnal motion around the Earth) motion with a period of one year, and a second motion with a period that is specific to each planet. For Mars, Jupiter and Saturn, the main orbit has a long period, and the epicycle has a period of one year.  For Mercury and Venus, the main orbit has a period of a year (keeping them aligned with the sun) and the epicycle has a period of less than a year.

When the right ascension of the planet is decreasing, then we say it's motion is retrograde. Retrograde periods occur during inferior conjunctions of the planet with the sun, and so the planet cannot be easily observed while it is retrograde.


Are any galaxies partially in the observable universe? Are there any galaxies that are partially with the visible universe? It seems unlikely as most of the universe is empty space however our circumference makes it seem likely as well. It would be a rather interesting thing to know about.

Perhaps, on rare occasions this will occur.  Quoting from Phil Plait in an  astronomycast ,

  Dr. Plait:    Yeah, because the universe is expanding ever faster every
  day, basically the distance we will be able to see, away from the
  earth, is going to get closer and closer to us. Anything past the
  distance will be moving away from us too rapidly to see and so
  eventually, all that is going to be left is some sort of super galaxy
  that is us and the Andromeda galaxy and all of the little dwarf
  galaxy’s, will have all combined into, sort of, one thing. Of course,
  all of the stars will be dead and then that’s it.

  So, you know, if you are a civilization, somehow, that evolves a
  hundred trillion years from now, the universe is going to look way
  different than it does now. You are not going to see any other
  galaxies; it’s just going to be you.

  Frazier:  You are not going to see any stars, because it will all be
  dark.

  Dr. Plait:    That’s right and for a long period of time that is all
  there is going to be, it’s just going to be that. Even on the scale of
  what we are talking about, we are talking about a long period of time,
  now instead of an octillion years, which is what, ten to the twenty
  seven, now we are starting to talk about ten to the fifty and sixty
  years, so far into the future, that the present age of the universe,
  13 billion years, is so small, that I don’t even have, ten to the
  sixty divided by ten to the thirteen is ten to the forty seven, that
  is such a huge number, that there is no ratio that exists today.

On those expansion scales, a galaxy is small (!!) enough that it'll pretty much disappear all at once.


Can radioactive decay rates be confirmed by looking at supernova? Radiometric dating relies on accurate knowledge of the half-lives of radioactive elements. Skeptics have argued that since many elements' half-lives are too long for anyone to have observed them, we can't know what they were in the distant past.

There are various counterarguments based on terrestrial evidence, like agreement between elements in the same sample, and corroboration from varves, tree rings, and ice layers.

Another counterargument I've seen is by looking at supernova, we get a window "back in time" to when their light began its long journey to us, and that much of this light was generated by radioactive decay, confirming that decay rates were the same then as now.

I don't quite follow this explanation, and would like some help. Distant starlight doesn't give us a rock sample in which to analyze proportions of parent and daughter elements. Does it still somehow give evidence of specific radioactive decay rates? If so, how?

I know next to nothing about astronomy, so please explain this to me like I'm a child.

I think I understand this now. I was trying to imagine how you could take a snapshot of a supernova and infer from it the half-life some some element. In fact, I think the measurement comes from watching the radioactive decay happen over time, in the aftermath of a supernova.
Openstax Astronomy 23.3 "Supernova Observations" shows a graph entitled "Figure 23.13 Change in the Brightness of SN 1987A over Time." It shows brightness increasing during the days right after the supernova, and the rest of the curve not following a consistent fading pattern. It explains:

One of the elements formed in a supernova explosion is radioactive nickel, with an atomic mass of 56 (that is, the total number of protons plus neutrons in its nucleus is 56). Nickel-56 is unstable and changes spontaneously (with a half-life of about 6 days) to cobalt-56. (Recall that a half-life is the time it takes for half the nuclei in a sample to undergo radioactive decay.) Cobalt-56 in turn decays with a half-life of about 77 days to iron-56, which is stable. Energetic gamma rays are emitted when these radioactive nuclei decay. Those gamma rays then serve as a new source of energy for the expanding layers of the supernova. The gamma rays are absorbed in the overlying gas and re-emitted at visible wavelengths, keeping the remains of the star bright.
As you can see in Figure 23.13, astronomers did observe brightening due to radioactive nuclei in the first few months following the supernova’s outburst and then saw the extra light die away as more and more of the radioactive nuclei decayed to stable iron. The gamma-ray heating was responsible for virtually all of the radiation detected from SN 1987A after day 40. Some gamma rays also escaped directly without being absorbed. These were detected by Earth-orbiting telescopes at the wavelengths expected for the decay of radioactive nickel and cobalt, clearly confirming our understanding that new elements were indeed formed in the crucible of the supernova.

So if I understand correctly, the logic would be:

SN 1987A occurred 168,000 light years from earth
Therefore, humans observed it 168,000 years after it happened
As they watched the supernova unfold, astronomers saw radiation that appeared to be generated by the decay of known, short-lived isotopes, and that decay happened at the same rates we observe today
Therefore, the decay rates of these isotopes were the same 168,000 years ago as they are today


What spectral type of star has an absolute magnitude of exactly 0? We know that Vega is the star that serves as the zero point for the UBV color scale, and has an apparent magnitude of nearly zero (+0.02). But its absolute magnitude is +0.58, making it rather far from $M=0$. So what spectral type would have an absolute magnitude of $0 \pm0.02$, and are there any stars that satisfy this?

There isn't a one-to-one relationship between spectral type and absolute magnitude.
Instead, there is a mean relationship with a fair bit of scatter around it. The reason is that the luminosity of a star of a given effective temperature depends on its composition/metallicity and how far along in its main sequence lifetime it is.
Basically, late B-type main sequence stars (say B7/B8V) have an absolute magnitude of about zero. Alternatively there are low mass stars ascending the hydrogen shell burning giant branch (types of about K2-K5 III) that would have an absolute magnitude of zero.