If the speed at which the universe contracts in reverse time decreases over time, then how did scientists arrive at a date for "The Big Bang"? If the speed at which the universe expands is constantly increasing (just like a derivative), then the opposite must also be true: the speed at which the universe contracts in reverse time is constantly decreasing, but, never reaches zero (like an asymptote).

If there is no exact point at which the speed of the contraction of the universe (in reverse time) reaches zero (because an asymptote goes on infinitely), then how did scientists arrive at a "date" for the "big bang"?

Since the speed of the expansion of the universe will go on and on ad infinitum, how can one arrive at a specific beginning point of an infinite function?

Under the assumptions of large-scale homogeneity and isotropy, cosmic evolution can be described by the scale factor $a(t)$ that describes by what ratio the large-scale structure of the universe has expanded or contracted. The scale factor is unitless, and by convention taken to be $1$ in the present epoch, and the rate of change of $a$ can be denoted $\dot{a}$, so the the universe is expanding whenever $\dot{a}>0$ and contracting whenever $\dot{a}<0$. The existence of a past singularity is then the claim that $a\to 0$ for some finite amount of time in the past.

The scale factor is related to the Hubble parameter by $H \equiv \dot{a}/a$. The Hubble parameter is very frequently and erroneously also called the Hubble constant, which is probably the source of your confusion. It is true that if $H$ is taken to be constant, then trivially $a = a_0e^{Ht}$ is the only possible solution, which only asymptotically tends to $0$ in the infinite past. If spatially flat, as our universe is, this is the de Sitter solution describing a universe dominated by a cosmological constant and negligible matter or radiation.

But that's simply not our universe. In the past, matter was dominant (and prior to that, radiation was dominant), and the de Sitter model simply isn't applicable. Simply put, the Hubble constant is not constant, and so should properly be called Hubble parameter.

In general, the evolution of the scale factor or the Hubble parameter is described by the Friedmann equations:
$$\begin{eqnarray*}
H^2\equiv\left(\frac{\dot{a}}{a}\right)^2 &=& \frac{8\pi G}{3}\rho - \frac{kc^2}{a^2}\text{,}\\
\dot{H}+H^2\equiv\frac{\ddot{a}}{a} &=& -\frac{4\pi G}{3}\left(\rho+\frac{3p}{c^2}\right)\text{,}
\end{eqnarray*}$$
where $\ddot{a}$ denotes the second time-derivative of $a$, the cosmological constant (dark energy) is included in the density $\rho$ and pressure $p$ of the universe, and for our spatially flat universe, $k=0$.

You may also wish to look up the deceleration parameter $q\equiv -\ddot{a}a/\dot{a}^2$ (so defined because the universe was once thought to be decelerating). But the main point is that $H$ is in fact not a constant, so it is consistent for the scale factor to become $0$ a finite amount of time in the past. How much time depends on the changing density $\rho$ and pressure $p$, which is studied in detail in the standard ΛCDM model.


Which is the most early type star with a planet discovered by radial velocity method? We have discovered planets in  Beta_Pictoris(A6V)here and Fomalhaut alpha(A3v)here.
However they were discovered by direct images.

There should be an stellar type limit by RV?
What is the most early type ever found?

HD 103774 b, detected by radial velocity, orbits a star of spectral type F5V according to The Extrasolar Planet Encyclopeadia.

HD 102956 b, detected by radial velocity, orbits a spectral type A subgiant, the star of the "earliest" type harboring a planet detected by radial velocity, I could find (as of May 2014).

The entry for
Kepler-88 c around a spectral type B star (according to the catalog as of May 2014) appears to be incorrect, or encodes something else, the star should be classified as spectral type G6V.


How to convert $K_S$ magnitude to $M_K$ My question is how do I convert a magnitude $K_S$, from a system like 2MASS, to $M_K$, and what are the differences. A use of this $M_K$ can be seen on Delfosse et al. (2000) in the $\log(M/M\odot) - M_K$ can be found. In the end I want to be able to use a $K_S$, and other possible parameters to derive the mass of the star.

Delfosse et al. (2000) - Delfosse, X., Forveille, T., Ségransan, D., et al. 2000, A&A, 364, 217

Since this seems to be a kind of niche question I will answer and leave some comments for future souls with the same problem.

To whom it may interest that may be searching for the same problem.
The solution is the basic $M = m + 5( \log (parallax) + 1)$. Parallax in arcsec.

The reason I'm going with K instead of V is due to a more independent relationships with the stellar metallicity

Parallaxes can be from catalogues like Hipparcus or Gaia.

If anyone is searching and gets here with the goal of using $M_K$ in the $\log (M / M_\odot) - M_K$ relation from Delfosse et al. (2000) keep in mind that in the paper the $\log$ relation is in $\log_{10}$ !!

Delfosse et al. uses "Johnson-Cousins-CIT", for that I followed equation (12) on Carpenter, J. M. 2001, AJ, 121, 2851.

If you want the $\log g$ I recommend the relation in:

Bean, J. L., Sneden, C., Hauschildt, P. H., Johns-Krull, C. M. and Benedict, G. F. (2006), ‘Ac-
curate m dwarf metallicities from spectral synthesis: A critical test of model atmospheres’, The
Astrophysical Journal 652(2), 1604.


Why do some pulsar planets have designations with a capital letter at the end? The pulsar PSR B1257+12 has three planets. Most triple-planet system have designation foo b, foo c, and foo d. However, I noticed that the three planets of this pulsar (PSR B1257+12) have a designation of A, B, and C as well as b, c, and d. Why did this naming convention deviate from the standard?

I will just cite this from Wikipedia:

On their discovery, the planets were designated PSR 1257+12 A, B, and C, ordered by increasing distance. They were discovered before the convention that extrasolar planets receive designations consisting of the star's name followed by lower-case Roman letters starting from "b", in order of discovery, was established. However, they are listed under the latter convention on astronomical databases such as SIMBAD and the Extrasolar Planets Encyclopaedia, with A becoming b, B becoming c, and C becoming d.

In a nutshell, this is the older nomenclature still kept because of historical reasons.


Instantaneous gravity in equation of motion for ephemerides calculation The equation of motion used for the calculation of orbits of solar system objects (Eq(27) in https://ipnpr.jpl.nasa.gov/progress_report/42-196/196C.pdf ) is formulated in terms of instantaneous variables, that is the acceleration of an object is assumed to depend on the instantaneous positions, velocities and accelerations of all other objects in the solar system (this has been confirmed to me by the authors of this publication). Would this not mean that information travels infinitely fast and thus contradict Relativity?

Would this not mean that information travels infinitely fast and thus contradict Relativity?

Look more closely at equation 27 in the referenced document. I'll simplify this as
$$\boldsymbol{\mathrm a} = \left(\sum_{B \ne A}\frac{GM_B\,(\boldsymbol{\mathrm r}_B - \boldsymbol{\mathrm r}_A)}{r_{AB}^{\,3}} \Bigl(1 + \text{other terms}\Bigr)\right) + \text{another term} + \text{yet another term}$$
Assuming Newtonian gravity, the "other terms", "another term", and "yet another term" all vanish, simplifying the equation to
$$\boldsymbol{\mathrm a} = \left(\sum_{B \ne A}\frac{GM_B\,(\boldsymbol{\mathrm r}_B - \boldsymbol{\mathrm r}_A)}{r_{AB}^{\,3}} \Bigl(1 + 0\Bigr)\right) + \boldsymbol{\mathrm 0} + \boldsymbol{\mathrm 0}$$
or just
$$\boldsymbol{\mathrm a} = \sum_{B \ne A}\frac{GM_B\,(\boldsymbol{\mathrm r}_B - \boldsymbol{\mathrm r}_A)}{r_{AB}^{\,3}}$$
All of those extra terms in equation 27 in the referenced document are due to a linearization of general relativity assuming that distances are very large compared to the Sun's Schwarzschild radius and that velocities are very small compared to the speed of light.


Coordinates of known black holes in our Galaxy (Milky Way) There are some articles in wikipedia with the list of known (or candidates) black holes here and list of nearest black holes. Also there is a map of Milky Way.

My question is where can I get approximate position of the known black holes in our Galaxy (Milky Way) as Galactic coordinate system or maybe some other galaxy coordinates?

If I go from the list of black holes to the page of a particular object like this there is a link to SIMBAD database like this. This SIMBAD database has galactic coordinates in this format

Gal coord. (ep=J2000) : 209.9563 -06.5399 [ 100 12 0 ]

where 209.9563 is a Galactic longitude, which changes from 0 to 360 degrees. -06.5399 is Galactic latitude, which is always around zero (+/- 10 degrees), because our galaxy is almost flat. Good explanation for galactic coordinates is on this wikipedia page.

To put the object on 2D map like this only Galactic longitude and distance to the object is needed. For example, this is the map of all black holes which I could find in wikipedia in our Galaxy.


Vectorial construction of tidal forces - or why is it centripetal at low tide I am seeking to understand why the tidal force is pushing towards the center of the earth (centripetal), at a point that is making with the center of the earth a 90º angle to the moon-earth axis.

This is very well seen in the common picture (from the Wikipedia "Tidal force" article):

I remember a vectorial construction of that from the French edition of Scientific American ("Pour la Science"), in their August 2001 issue. But I lost it, and it is now behind a pay wall:

https://www.pourlascience.fr/sd/geosciences/les-caprices-des-marees-4441.php

I used to have the physics/mathematical background to understand that, and perhaps I could dig it out of the decade-old sediments of my memory. The vectorial construction looked interesting to me as slightly easier to explain to people around than the differential equations.

The short answer which may or may not be an "Aha!" answer is that what is plotted is what's left over after a much larger, uniform force is subtracted.

The uniform force is the Force from the Moon evaluated at the center of the Earth, and the arrows show the deviation of the actual force from that average.

Why do we do it that way? When looking at the tidal forces on the oceans we treat the Earth as a rigid body with spherical symmetry. With that, we can use a variation of Newton's Shell theorem to say that the extended Earth will move the same way as if it were a point mass at its center.

Now the oceans are fluid (the opposite of rigid) and each bit responds to the Moon's force locally.

That force is (skipping constants unnecessary to make the cartoon plot)

$$F = -\frac{\mathbf{\hat{r}}}{|r|^2}  = -\frac{\mathbf{r}}{|r|^3}$$

where the vector $\mathbf{r}$ is drawn from the Moon to some point on the Earth and $\mathbf{\hat{r}}$ is its unit vector. If the Earth's center is at $\mathbf{\hat{x}} R$ ($R$ is the Moon-Earth distance) and you subtract $-\mathbf{\hat{x}}/R^2$ you'll get that image.

In the plot below I've chosen the Earth-Moon distance to be only 10 Earth radii to highlight the slight left-right asymmetry. The tidal force is stronger on the side closer to the Moon.

import numpy as np
import matplotlib.pyplot as plt

R = 10.0
r_moon = np.array([R, 0], dtype=float)[:, None]
earth  = np.zeros(2)[:, None]
theta = np.linspace(0, 2*np.pi, 49)
positions = earth + np.array([f(theta) for f in (np.cos, np.sin)])

r = positions-r_moon
F = -r * ((r**2).sum(axis=0))**-1.5

r = earth-r_moon
Fmean = -r * ((r**2).sum(axis=0))**-1.5

Ftide = F - Fmean

if True:
    plt.figure()
    plt.subplot(2, 1, 1)
    (x, y), (Fx, Fy) = positions, 50.*Ftide
    plt.quiver(x, y, Fx, Fy, width=0.005)
    plt.plot(x, y, '-b')
    plt.xlim(-2, 2)
    plt.ylim(-1.5, 1.5)
    plt.gca().set_aspect('equal')
    plt.subplot(4, 1, 3)
    for thing in F:
        plt.plot(thing)
    plt.subplot(4, 1, 4)
    for thing in Ftide:
        plt.plot(thing)
    plt.show()


What will a future cold and "dead" Jupiter look like? I heard Jupiter is dying like getting smaller. For example, from Guillot et al. 2004 3: The Interior of Jupiter (also here):

(Jupiter)... is still contracting at the rate of ~3 cm per year while its interior cools by ~1 K per million year.

When will it completely die, how much smaller will it be then, and what will it look like?

As earlier answers indicate, Jupiter will slowly shrink until it is some 10-20% smaller with elements slowly settling depending on density and solubility.
One can calculate a rough Kelvin timescale for the ratio between binding energy and the blackbody power of the surface to estimate how quickly it cools down (this number will be nudged by the above considerations a bit, so it is an order of magnitude estimate):
$$\tau_{Kelvin}=\frac{3GM^2}{20\pi\sigma R^3 T^4}.$$
For Jupiter this timescale is 54 billion years.
As the universe cools eventually it becomes a ball of layered solid degenerate matter, with a frozen hydrogen crust. This will take longer than $\tau_{Kelvin}$ due to various mild heating processes like crystallization heating, tidal heating from the moons, possibly dark matter annihilation, and other low-energy processes that keep objects warmer than the CMB.


Mars is dark, surrounding is bright, what am I doing wrong? I just got a Celestron 8 and I'm a complete rookie: set up scope early so it would acclimate to the temperature. No difficulty locating Mars, but I see a dark planet with its circumference slightly illuminated and the night sky around it is bright? (scope in good dark location and nice dark sky) No amount of fine tuning makes a difference. What am I doing wrong?

Sounds like you are very out of focus. Dial the focuser all the way out towards you. The little knob on the right.


Historical lunar occultation of Venus seen from Sicily in the winter? I am looking for dates in history, from 400 years ago or older, when there was an occultation of Venus by the Moon during Winter as seen from Sicily.  Venus was rising, and according to the text that I am trying to fact check, it is not clear what phase the moon is in. The text from Shakespeare reads there were "nine changes to the watery star"  This is interpreted to mean the moon, as it effects the tides.  So the moon should be 9 days away from a full moon.

Or if an occultation is too rare would there be an appulse that matches the criteria?

1) Venus is rising.
2) The moon is waning nine days.
3) As seen from Sicily.

I am led to understand that this is a rare event, happening once every 200 years or more, and I would be curious to know about any other unusual astronomical events that occurred within +/- 23 or 24 years of this rare event.

Where could I find this information?  Is there a star chart I could check?

If the Moon is 9 days past full, then it can't be occulting (or even at its closest approach to) Venus.  Since Venus is an interior planet, it never gets far from the Sun in the sky as seen from Earth; its maximum elongation (maximum angle between Sun and Venus) is 47°.  So to occult Venus, the Moon would also have to be that close to the Sun in the sky.

However, when the Moon is 9 days past full, it is still more than 70° from the Sun in the sky (verified with calculation of lunar position and phase from JPL Horizons; the "SOT" angle given there is "Sun-Observer-Target" and thus the Sun-Moon angle as seen from Earth). The Moon doesn't get within 47° of the Sun until about 11 days past full.

On a more literary note, the passage "nine changes of the watery star" (from The Winter's Tale) seems to be typically interpreted in commentaries to mean nine months have passed, i.e. each "change" is a full cycle of lunar phases.

EDIT:  If you want to search for historical occultations in general, it looks like the Windows software Occult 4 should be able to do it.  The PDF guide has some tips for setting it up.


How long does it take Dawn to orbit Ceres? I haven't been able to find the orbit time of Dawn around Ceres.
Do you know how long it takes?

As of March 6, 2015, Dawn has entered orbit around Ceres.  But it's really so far only "captured by Ceres' gravitational pull".  Then, it was still 61,000 km from Ceres.

It's slowly spiraling down into an orbit that JPL's Dawn Journal calls "RC3", which will be 13,500 km above Ceres.  This orbit will last 15 days per revolution.

  It will take about 15 days to complete a single orbital revolution at this altitude.

The first news link above indicates that Dawn will reach RC3 on April 23, 2015.

This picture from the Dawn Journal gives us a picture of Dawn's path: it's not completing any revolutions around Ceres yet; it's still approaching the RC3 orbit as of this writing (March 10, 2015).

So as of now, March 10, 2015, there is no orbital period yet.  But when RC3 is achieved, it will be a slow orbit -- 15 days.

There are other orbital altitudes planned past RC3.

"second orbital phase, survey orbit" - Altitude 4,400 km, Orbital period 3 days, 3 hours
"high altitude mapping orbit" - Altitude 1,470 km, Orbital period 19 hours
"low altitude mapping orbit" - Altitude 375 km, Orbital period 5.5 hours

According to Dawn's schedule, the "survey orbit" will only last 22 days before it spirals in to the "high altitude mapping orbit" in August 2015, although it's unclear exactly when the "survey orbit" will take place.  The "low altitude mapping orbit" will be in November 2015.  In between orbits, when spiraling down to the next scheduled orbit, Dawn's orbital period will slowly decrease.  At the last orbit, Dawn will be left in this orbit after its batteries and hydrazine fuel are exhausted.

Update (Oct 23, 2015)

Excerpt from latest news update from the Dawn mission (Sept 30, 2015):

  Dawn is currently orbiting Ceres at an altitude of 915 miles (1,470
  kilometers), and the spacecraft will image the entire surface of the
  dwarf planet up to six times in this phase of the mission. Each
  imaging cycle takes 11 days.

  Starting in October and continuing into December, Dawn will descend to
  its lowest and final orbit, an altitude of 230 miles (375 kilometers).
  The spacecraft will continue imaging Ceres and taking other data at
  higher resolutions than ever before at this last orbit. It will remain
  operational at least through mid-2016.


When a star reaches the red giant phase, why does it become more opaque? Please refer to this answer from Quora:

  ... a star will become a red giant before it begins burning helium. In fact, it will bloat into a red giant while still burning hydrogen in a shell on the surface of the helium core. Shell burning, however, does release more energy than core-burning, but even that alone wouldn’t cause the star to become a red giant, since it could just shine brighter. The real culprit is the combination of the higher energy production rate AND that the star has a higher opacity at that stage in life. This produces an energy crisis where the energy cannot escape radiatively fast enough, and convection would have to be supersonic to rebalance the star. Since supersonic convection is highly disfavored (i.e. impossible), the star greatly expands to the point that the energy flux at the burning shell matches the energy flux at the now much larger surface again.

This is one of the best explanations I've found for how stars transition to red giants. However, the section I've bolded is confusing to me: why would the star become more opaque, and what causes that to occur?

There are some answers in this lecture and this one although even there the author admits that the whole story is complicated, and not perfectly known.

It seems that shell burning and contraction of the no-longer-burning core of the star causes some expansion of the outer layers (this is described as the "mirror principle" and partly explained in lecture 12 (section 12.4). This causes the outer layers to cool, until H- ions start to form in them (around 5000K). These interact with light much more than neutral hydrogen atoms or bare protons (H+) which accounts for the increase in opacity.


Point Spread Function size: Full Width Half Maximum (FWHM) vs Sigma I having been studying some astronomy papers related to galaxy observations and I realized that every time they want to express the size of the Point Spread Function (PSF) of a system which can be approximated by Gaussian they use the Full Width Half Maximum (FWHM) and not the sigma. Is there a reason behind that?

Thank you.

FWHM is the indicator of the width of a Gaussian that is easiest to measure, and is least error prone in terms of actual physical measurement (due to the slope of a Gaussian being the highest near half maximum - not exactly sure about this, but it does seem like that's the case - so error in y contributes least in the measurement of width, relatively. If this is not clear, look at a noisy Gaussian - something like this: http://www.astronomie-amateur.fr/Documents_Supernovae/Mesure_Vexp_gauss.PNG - and you'll know what I mean. It seems most sensible to estimate width using FWHM). And it has a simple correlation to the sigma with a factor of 2.35 if the distribution is actually Gaussian (which is often the assumption for PSFs - correct me if I'm wrong). Measuring actual standard deviation of data with errors is much more complicated than just simply measuring FWHM, and often it is enough to know the FWHM estimate. Hope this is the answer you were looking for.


What is meant by Electronic Beam Steering and how is it achieved? I recently came across an article about the Murchison Wide Field Array which said directivity is achieved by Electronic Beam Steering and not by Mechanical methods. It would be helpful if someone can tell me: how do they achieve it?

This can actually mean a couple of different things, but in the case of steering an MWA tile, Jeremy is quite correct. Each tile has a 'beamformer' box that all of the individual antenna cables plug into. This box physically delays each signal appropriately (by having a longer path to go through for more delay) based on the signals sent to it for each pointing. Because of the quantised path lengths there are what are called 'sweet spots' that are used, in other words arbitrary pointings don't normally happen you just dial in the closest 'sweet spot' to where you'd like to point.

There are also version that do this in software (and in fact in a special phased mode the MWA does this between tiles sometimes), where 'turns of phase' are used instead but it is essentially the same idea.


How much could Earth's orbit shift without causing an ecological catastrophe? Seeing the news about A/2017 U1 made me wonder what would happen if a similar object, but with a mass close to Earth's, had a close call with Earth. Could it shift Earth's orbit enough to quickly and significantly change the conditions for life?

This answer sort of gets it it, but it's not clear how much of an effect the change in orbit would have.

Depending on how one defines "quickly" and "significantly," I think a gravitational influence strong enough to do that would tear apart the Earth through tidal effects long before the change in orbit would make the planet unsuitable for life.


Does the magnetic field of earth restore? The magnetic field of earth protects us from solar wind as we know it, but after the solar wind interacts our magnetic field.

Does the magnetic field of earth restore after it deflects the solar wind from the sun?

Remember that a magnetic field is just that:  a field . There's no mass there.   Its shape depends on the electromagnetic properties of Earth -- primarily the behavior of the outer core.
When there's a significant Solar ejection,  the  electrically active particles in that mass cause local changes in the magnetic field pattern (equivalent to isobars , or lines of constant field strength, etc.)   So long as the solar ejecta do not damage the Earth in any significant way, once the particles are gone, the field lines will revert to their default layout.


How much magnetic attraction is there between the Sun and Earth? How much magnetism is there between the Earth and Sun? Would the distance from the Sun differ for not the Earth iron core and how much?

Warning: I am only rather mediocre on astronomy. It was a hobby of mine when I was in second grade to now, so I pretty much know a decent amount of info. But my speculation for this question may be some distance off.

First of all, let us look at the Sun and Earth. The Earth has a [25~65 microtesla] magnetic field due to the motion of convection currents its molten [mostly iron and nickel] outer core. These convection currents are caused by heat escaping from the solid inner core. Heat from below heats up the liquid, which rises, cools, sinks, heats up, rises... you get the point. These convection cells creates charge/electricity in the outer core. Now, when you hear electricity, and then, iron/metals, you will obviously know that iron would conduct electricity. Sort of like an electric magnet nowadays. That is exactly so in Earth's outer core. The electricity generated by the convection cells is [what I believe] conducted by the outer core and amplified/extended to form the magnetic field. This is the Dynamo theory, which contains three parts:

  There are three requisites for a dynamo to operate:

  1) An electrically conductive fluid medium

  2) Kinetic energy provided by planetary rotation

  3) An internal energy source to drive convective motions within the fluid.

  In the case of the Earth, the magnetic field is induced and constantly maintained by the convection of liquid iron in the outer core. A requirement for the induction of field is a rotating fluid. Rotation in the outer core is supplied by the Coriolis effect caused by the rotation of the Earth.

The sun's [100 microtesla] magnetic field compared to Earth's is a really large and complicated. Since the sun is THAT LARGE :P, its magnetic field extends beyond the planets forming the heliosphere, a protection against cosmic rays from interstellar space. Unlike but similar to the Earth, the magnetic field of the sun is derived from the electricity conductive plasma inside the sun, but also from convection. Since the sun is not a stable ball, but rather a giant ball of fluid in a medium, it's rotation speeds on the equator and the poles are different, and this bends/stretches the magnetic field, creating sunspots, CMEs, Solar Flares, and many other "space weather" events.

To answer what you are asking, the magnetic attraction between Earth and the Sun should be rather strong, but not to an extreme extent. If you were to take 2 magnets of the same magnetic strength ratio, the stronger magnet the readily attract the weaker magnet. This same affect should be that of what the Sun is sort of doing. But even along with gravity, the Sun's pull on Earth cannot force it into a collision course due to Earth's sideways momentum. As commented on by uhoh, the interplanetary magnetic field is the magnetic fields of all objects in the Solar system. The sun, being the strongest one, takes dominance. But what is rather confusing is that there is a thing called Solar Wind. The Solar Wind is a rather "pushing" stream of charged that will "push" away particles if the objects doesn't have a magnetic field to repel that force. The Solar Wind carries the Sun's own magnetic fields away from it, but since magnetic lines cannot cross celestrial objects, they "drape" themselves onto the magnetic field-less object and form the "magnetosphere" of that object. I would imagine this would have little effect however on the attraction of the Sun and that object(earth, asteroid...) itself.

If you are looking for numbers, a hefty equation to calculate the electrical
magnetism oh an object is the Boit-Savart Law.

  If you have a
  current $I$ flowing through an infinitesimal length
  $\mathrm{d}\vec{l}$ then the field at a point $\vec{r}$ is given by
  \begin{equation} \mathrm{d}\vec{B} = \frac{\mu_0}{4\pi} \frac{I
> \;\mathrm{d}\vec{l}\times\vec{r}}{|\vec{r}|^3} \end{equation} The
  field does still drop of with the inverse square of the distance, but
  the vector nature of the current makes the formula rather more
  complicated. Generally this expression has to be integrated over the
  length of a wire carrying the current. This can get quite complicated
  and so this method is generally only used for simple situations such
  as the field around a magnetic dipole.  Having found the field, the
  force on a infinitesimal length of current can be found from.
  \begin{equation} \vec{F} =
> I\;\mathrm{d}\vec{l}\times\vec{B}\end{equation} This is again
  analogous to the electrostatic case but with added complication due to
  current being a vector.

    https://physics.stackexchange.com/questions/122258/what-is-equation-to-find-force-of-magnetic-attraction

This can be used to calculate the force of Earth's magnetic field, and the Sun's magnetic field, since they both are electrically generated. The resulting numbers can be put into the magnetic attraction equation:

$$F=\frac{\mu q_{m1} q_{m2}}{4\pi r^4}$$
where

$F$ is force attraction in newtons.
$q_{m1}$ and $q_{m2}$ is the magnitudes of magnetic poles in ampere-meters.
$\mu$ is the permeability of the intervening medium is tesla meter per ampere, henry per meter or newton per ampere squared.
$r$ is the separation in meters.

Thus, the attraction between the Earth and the Sun can be calculated in the following means:
$$\implies \operatorname{B}(\text{Earth})=q_{m1}=7.84027\times 10^{22}$$
$$\implies \operatorname{B}(\text{Sun})=q_{m2}=3.48526\times 10^{29}$$
$$\implies \mu\approx 10^{-6}$$
$$\implies r=\operatorname{AU}=149597870700$$
$$\implies F=\frac{10^{-6}\cdot (7.84427\times 10^{22})\cdot (3.48526\times 10^{29})}{4\pi\cdot 149597870700^4}$$
$$\implies F=\boxed{4.34165\text{ newtons}}$$
If the Earth doesn't have a liquid, conductive outer core, it wouldn't have a magnetic field. This would be similar to Venus, which has a super duper weak magnetic tail and magnetosphere from the Solar Wind(suprising is that it still retains an atmosphere. I will ask why...). However, we can use a real world example of such. Let's take Mars as the example. Mars used to have a magnetic field from its liquid outer core 4 bya, but this field was destroyed, either from repeated bombardment from large astroids which would have disrupted its internal structure, or from the cooling of its liquid core. Thru recent evidence from tests on Mars' crustal fields, it has shown that Mars' earlier magnetic field is more powerful that of Earth's. Now all that's left is some umbrella shaped fields concentrated on it's southern hemisphere. This offers little protection and would sufficiently answer your second question. You asked if not for the conductive, LIQUID, outer core, what would happen to the distance as possibly the attraction between the Sun diminishes and Earth wanders more outwards. I would suggest not such. Mars would fit your speculation, but its 4 bya orbit and current orbit has minute difference. I would postulate that the magnetic field degration would have some effect, but such effect would be rather unnoticeable, because even though electromagnetism is a very strong force, realize that gravity is the strongest force in the universe(black holes). The attraction between the celestrial bodies orbiting the Sun and the Sun itself is almost completely gravity.

I may be off, or way wrong, but to what I know, I believe this is more or less the right answer.


Are there any works discussing planetary bodies being forms of life? I'm searching for works or individuals that consider and discuss the idea of planetary bodies being considered stand alone living creatures, rather than objects either capable or incapable of supporting life.  Anyone familiar with something along those lines?

Look up Gaia hypothesis, James Lovelock, and Lynn Margulis.


What qualifies as a good place for stargazing, i.e. with least light pollution? I know that minimal light pollution is a must for stargazing, and a place which is away from civilization is better.

Does altitude of a place matter for light pollution?

Does it affect the quality of star gazing?

A good question, and in the early 2000s John Bortle published a categorization of a variety of conditions, with descriptions for each category. It is the commonly used scale to describe to others the sort of conditions at a location.

Probably one of the more significant factors provided by a dark sky site is: how faint do stars have to be for you not to see them anymore (overwhelmed by light pollution). There are specific stellar regions, each a triangle, and the idea is you could how many stars you can see in the triangle, and you look the number up in a table, and it will tell you the magnitude limit you are perceiving at that site.

Yes, high altitudes also help - less air between you and the stars means better seeing conditions/less atmospheric distortion.

Being able to see more stars IS better for star gazing. Living in a city, I am continually frustrated at the poor skies, and love being in dark sky locations like the International Dark Skies Reserve at Tekapo, New Zealand.


Does Polaris have a trail in the sky timelapse? I came across few sources that claim Polaris movement cannot be observed with the naked eye. But then, I saw images of the star trails (timelapse), and they clearly show Polaris not quite at the true north and moving around the Earth's axis with a radius that is very visible. See for instance the following images:

However, this image from Wikipedia does not show a trail for Polaris:

What am I missing here?

Does Polaris have a trail in the sky timelapse? Yes. Every star has a trail, but it's smaller if it's closer to the pole. Polaris is close to the pole but it's still 45 (arc)minutes away. I don't think it's visible with the naked eye; no star trail is visible with the naked eye as you only see one 'instance' of the sky. What perhaps would work, and is more feasible for stars closer to the celestial equator, is trying to see the movement of a star compared to an object on Earth, e.g. a tree, a building or the horizon.

The exposure times of the photos are different, as @J.Chomel notes in the comments. You can calculate the exposure time by looking at the arc length; in 23 hours and 56 minutes (a 'sidereal day'), a star will make a full circle.

The trails on the last photo have an arc length of about 8 degrees, corresponding to about 30 minutes; the first two photos are more like 9 resp. 5 hours. Also, the zoom factor (of the lens) may play a role; if a photo is zoomed in more, Polaris' movement is more likely to be visible. I'm not sure if this plays a role in this photos, though.


How to differentiate between images of a gravitationally lensed object The light coming from a distant galaxy or star can be distorted en route to Earth; this distortion is caused by the curvature of space when a massive object is present. My question is, how can we tell whether these doppelgangers are from the same galaxy (especially in the case of a star)?

In the case of multiple images of a background, distant object, the answer is relatively simple. You take a spectrum of the multiple images or parts of an extended lensed image and you see whether the spectrum looks the same, and in particular whether the redshift of the multiple images are the same.

Gravitational lensing affects light of all wavelengths equally, so regardless of the path taken, the spectrum should be unaltered, except that the different paths take different amounts of time to travel from the original object to us. So, if the background source, or its spectrum, are time-varying, then the lensed images could also appear different. However, distant galaxies do not change their redshift on such short timescales and so this should still be the same regardless.


If months are based on the moon, then why are the months longer in the Gregorian calendar than lunation? What I have always thought / known is that:
Days are based on the period of rotation of the earth.
Months are based on the moon.
Year is the time taken by the earth to make one revolution around the sun.
But if we take a look at the lunation it takes on average 29 days, 12 hours, 44 minutes, and 28 seconds. Why is it that the months then in the Gregorian calendar have 28-29 / 30 / 31 days?

Ancient Egyptians and Mesopotamians came up with “administrative” calendars of 30 days, that were easier to calculate than “real” lunar months of sometimes 29, sometimes 30 days. At the end of the 12 months, they added five days of religious observances, sometimes positive, sometimes negative.
When Julius Caesar conquered Egypt, he knew about the very regular Egyptian calendar and appointed his mathematicians and astronomers (if we can really call them that, as Roman science wasn’t very developed) to discuss with their Egyptian peers and devise a better calendar for the Roman Republic (reminder that Caesar was not the Emperor; the first one was Augustus, his [adoptive] son, and not right away).
Eventually, it was decided in Rome that, in order to follow with the seasons, the already-existing Republican Roman calendar was to be modified. It contained 10 months of 29 and 30 days plus a “winter” of 59 days, which became divided in January and February. As Julius Caesar was the one who brought this change, he was “honored” by having a month renamed from Quintilis (“Fifth”) to Julius (our July)—and that month was given 31 days, as well as a few others. Also, the beginning of the year was shifted (at this point or later; it’s not clear) to January, making the “numbered months” off with their new rank—e.g. “October” is not eighth (“octo”) anymore.
Caesar’s reform was not well applied, and eventually, Augustus had to enforce the new calendar, and he was also “honored” with a 31-day month, right after Julius’, so that’s how Sextilis (“Sixth”; and it had 30 days) became Augustus (with 31 days).
The rest is basically adjusting here and there to make it a total of 365 days at the end of the year, someone sometime having the “brilliant” idea of alternating 30- and 31-day months (a little like nature “alternates” between 29- and 30-day lunar months).
Side note… In some languages, the word for Moon and the word for month is the same—it’s already quite similar in English! For example, Romanian has “Luna” for both, and Korean has ”dal.”


Saw a bright orange object in the sky. What was it? Some friends and I were backpacking in the mountains, and at night we saw a very bright orange object in the sky. It looked brighter than Venus. After a couple minutes we realized it was moving horizontally very slowly. Its movement wasn't noticeable unless looking at it next to a reference point like a tree. This was way brighter and slower than any satellite I've seen. Anyone know what this was?

If the bright orange object was low in the southeast around midnight, it may have been Mars.
In late July and early August 2018, it will be only 0.39 AU away from Earth and shine with apparent magnitude -2.7, slightly brighter than Jupiter.
This is not as bright as Venus but may stand out more strongly in a midnight sky than Venus does in twilight.
For most of Mars's 2.1 year synodic period, it is farther from Earth and appears fainter.

As the Earth rotates, celestial objects appear to move about 1/4 degree per minute from east to west.
For observers in the northern hemisphere, objects near the southern horizon appear to move horizontally left to right.


What could be a star-like object that shines at daytime for few seconds? What's the probability to see it? Some years ago I looked up to the blue sky, around 2 PM, and I saw, for 2 or 3 seconds, something like a star fading out.
Could it be a Supernova or another phenomena? Is it highly possible?

Several satellites (but most notably Iridium) have large reflective panels.  If aligned with the sun properly, it can shine with sufficient brightness to be visible in the daytime.

An Iridium Flare usually lasts only a few seconds.  It sounds quite consistent with your description.

It can be hard to tell in a blue sky with no nearby references, but an Iridium satellite would also be moving while fading out.

There are sites that can show you upcoming flares for your area.  If you knew the exact date in the past, they could probably even "post-dict" what flares were available then and see if it matched.


Where are heavier elements than iron produced? Specifically in the milky way, if that changes anything.

I feel like it should be in the disk, since stars there are richer in heavier elements, but I'm not sure.

The question of "where" the heavy elements are produced is slightly ambiguous. I gather from your mention of the Milky Way that you are interested in "where in the Milky Way" are the heavy elements produced, rather than what type of stars are they produced in.

However, to answer the former, you need to understand the latter. So I point you to my answer on Physics SE for a summary of that.

The heavy elements can be divided in a number of ways. Most are produced by neutron capture in either the slow s-process or the more rapid r-process. Different elements are produced in different quantities by these two mechanisms and are created in different types of star. s-process elements are created inside massive stars before they become supernovae, but also in stars of more modest mass that never go through the supernova stage. The r-process is now thought to occur predominantly in neutron star mergers, but there is probably also a contribution from type II supernovae, particularly for the lighter r-process elements. A summary, which still has plenty of uncertainties, but probably gives a reasonable summary of current thing is shown below (diagram produced by Jennifer Johnson).

Your question then amounts to where these stars were when they created these heavy elements towards the ends of their lives, and as a function of time since the Milky Way was born. The key facts here are that the progenitors of type II supernovae are massive stars with short lives. They are born in gas-rich regions and do not have time to travel far from this location before they explode. Thus most of the elements they produce are produced in the bulge and disc regions of the Milky Way. This is especially true for the bulge, which seems to have formed quite quickly at the start of the life of the Milky Way and there should (and indeed does seem to be) a radial gradient in metallicity increasing towards the disc plane and towards the Galactic centre.

Type Ia supernovae result from exploding white dwarfs can produce elemts at the iron-peak and somewhat beyond, mostly by "standard" fusion than by neutron capture. These events may have been produced by comparatively lower mass and long-lived progenitors, and may also have had long lives as white dwarfs before the supernova event. This means that type Ia supernova progenitors are spread more widely. They have had plenty of time to migrate from where they were actually born, so we expect their heavy element production to be more dispersed - still concentrated towards the disc, but much less so than the progenitors of type II supernovae.

The progenitors of the asymptotic giant branch stars that can produce lots of s-process elements can also be quite old. The elements produced by these stars would be distributed in a similar way in space and time to those produced by type Ia supernovae.

Finally we have merging neutron stars. These are quite problematic and not much is known about their progenitors or evolutionary paths and histories. They must originate from massive stars, with short lives that are born close to the Galactic plane or bulge. However, after both stars in the systems have exploded as supernovae and left behind neutron star remnants, the final neutron star merger can take place much later, due to a gradual inspiral triggered by the emission of gravitational waves. This gives the neutron star pair a long time (hundreds of millions or billions of years) to move about in the Galaxy before the merger takes place. In addition, the supernova explosions can give a "kick" to the neutron stars that can displace them well out of the Galactic plane with fast velocities. I expect therefore that the products of neutron star mergers could have been produced over a very large volume in our Galaxy.


Can a single impact event trigger a runaway greenhouse? Somewhat related to this question: Is the Earth going to evolve towards Mars' fate or Venus' fate?
Human civilization may be fragile, but life itself is extremely resilient, with extremophiles and microscopic lifeforms able to withstand such harsh conditions that make the end of all life possible, but very unlikely. Once in a while articles are published, like this, listing a large enough impact event as one possible scenario, besides supernovas and gamma-ray bursts. The threshold for large enough they adopted was one with sufficient energy to boil all oceans, requiring about 5.6 * 10^26 J of energy and a impactor with a mass around 1.7 * 10^18 kg. Only 19 asteroids fit the bill, counting Vesta and Pallas.
But water vapour is known to be a strong greenhouse gas. So, couldn't be the case, instead of boiling whole oceans, a single impact could release just enough water steam to kickstart a runaway greenhouse like the one in Venus? I imagine that would happen if a substantial fraction of the atmosphere becomes steam (guess 50%). We can estimate the mass of Earth atmosphere by multiplying atmospheric pressure by planet surface area, giving about 5.2 * 10^18 kg. We can estimate the energy necessary to release a equal amount of steam, multiplying this mass by water Enthalpy_of_vaporization, and the impactor kinetic energy required, by E = mv^2/2(see below).  I assumed a 20km/sec speed that I read somewhere to be close to average for a collision with Earth (the range goes from about  11km/sec to 70 km/sec).
This gives us E ~ 1.16 * 10^25J and m ~ 5.82 * 10^16 kg for impactor energy and mass, smaller than the ones given in the link above by one and two orders of magnitude, respectively. That bring us to Hale-Bopp size range.
I'm aware Earth withstood multiple large impacts, in Late Heavy Bombardment and before, and yet the water just condensed back and so we had no runaway greenhouse induced by released steam. But as the Sun gets older and hotter, will that always be the case? Perhaps the dust released in the impacts negates the effects of water vapour in the atmosphere, but I can imagine some scenarios where the ratio of dust to vapour is minimized, like a impact in deep ocean, for example. What are the odds of something like this happened to Venus, flipping a borderline-stable Earth-like world into the runaway greenhouse we see today? Could be the case Earth is already past the same threshold, just waiting for a suitable impact to flip the switch?
I found a paper(1) discussing a sudden change from runaway icehouse to runaway greenhouse, but nothing on impact-induced transitions.
Calculations:
Python 3.6.9 (default, Jul 17 2020, 12:50:27)
[GCC 8.4.0] on linux
Type "help", "copyright", "credits" or "license()" for more information.
>>> m_impactor = lambda E, v: 2*E/v**2 ## From kinetic energy formula
>>> E = lambda m_water: 2.23e6*m_water ## 2.23 MJ/kg^3 = approximate heat of vaporization for water
>>> import math
>>> m_water = lambda P, g, R: 4*math.pi*P*R**2/g ## P = atmospheric pressure (about 101 kPa); g = gravitational acceleration (9,8m/s^2); R = Earth radius (about 6350 kilometers).
>>> m_impactor(E(m_water(1.01e5, 9.8, 6.35e6)), 20000)
5.822741097812724e+16
>>> E(m_water(1.01e5, 9.8, 6.35e6))
1.1645482195625448e+25
>>> m_water(1.01e5, 9.8, 6.35e6)
5.222189325392578e+18

References.

Yang, Jun, et al. “Abrupt Climate Transition of Icy Worlds from Snowball to Moist or Runaway Greenhouse”. Nature Geoscience, vol. 10, no 8, agosto de 2017, p. 556–60. www.nature.com, doi:10.1038/ngeo2994.

Sloan, David, et al. “The Resilience of Life to Astrophysical Events”. Scientific Reports, vol. 7, no 1, julho de 2017, p. 1–5. www.nature.com, doi:10.1038/s41598-017-05796-x.

What your calculations don't account for is the dynamics of the meterological system.  An atmosphere with $10^{18}$kg of water vapour would be a powerful greenhouse, but at the temperatures on Earth, there is no way that the atmosphere could hold so much water vapour, it would rapidly condense.
Instead, in the case of a major impact into the oceans, a lot of water vapour is produced, which rapidly condenses to clouds because the Earth isn't hot yet.  The clouds will tend to cool the Earth (by raising albedo).
The climatic effects will be significant, but you would expect all the water vapour generated to condense fairly rapidly, and far faster than the greenhouse effect of the water vapour.
The Earth has certainly experienced such impacts into the oceans, (this is a Chicixlub level event, expect once every 100 million years or so, and ⅔ will fall into the ocean) and yet no runaway greenhouse has occurred.


Are epicycles and mean Earth/Sun actually useful astrophysical concepts? Copernicus put the Sun in the center of the Solar system. This turns the Moon's orbit into an epicycle around Earth orbit. And the Jovian moons (later discovered) have orbits which are epicycles around Jupiter. Since Copernicus created a physical epicycle in the Moon, it maybe wasn't so strange to think about "epicycles of epicycles" as he did to improve the model's fit to calendar data.

Actually, Ptolemy and Brahe did not put Earth in the center of their Solar system models. They used a fixed point in empty space near the Earth which they called "mean Earth". And Copernicus used a corresponding "mean Sun". And as a matter of fact, planets do turn around a barycenter which is not in the center of the Sun, but nearby.

Do epicycles and "mean Earth/Sun" actually have physical correspondents in the orbits of moons and the barycenter respectively? That'd be a bit funny since they are purely geometric constructions with the purpose to make data fit better, created even without a thought about gravity or anything physical at all.

I'll say yes here, but only if the epicycles are used as approximations. Here's how they would be useful:

The eccentricity of the celestial body's orbit is low, i.e. $e<<0$. This means that it would be nearly circular, and epicycles could therefore be reasonably accurate. For example, it would not be a good idea to use epicycles to model Mercury's motion, because it has a high eccentricity. Comets would also be poor choices. Triton, on the other hand, has low eccentricity.
Only a vague approximation is needed. If you aren't looking for calculations with a high degree of accuracy, you could use epicycles. It would be a bad idea to use epicycles to put a spacecraft into a precision orbit, but a better idea to use them in a toy simulator. NASA, however, has higher standards.
You don't need to use general relativity. One of general relativity's most important predictions was the precession of Mercury's orbit. Over long periods of time, general relativity triumphs over classical theories when it comes to this accuracy. It would be hard to use epicycles to model this - although potentially possible.


Star visibility above horizon map likewise day & light map or moonlight world map I am looking for an option to present the given star visibility world map likewise day & night map or moonlight world map like below:

Question: Is it possible to generate a map like this for some particular star or at least the point in the sky defined by celestial coordinates?

As long as we can live with a few tens of kilometers accuracy, we can treat the Earth's surface as a sphere.
The subpoint of the object (the point on Earth at which the object appears at the zenith, overhead, or the central point on Earth seen from the view of the object if they could see Earth clearly) is simply given by
$$\text{lat} = \text{Dec}$$
$$\text{lon} = \text{R.A.} - \text{GST}$$
where GST is Greenwich Sidereal time
To make your plot we draw a great circle around the subpoint, then make a simple Equirectangular projection or lat/lon projection.
In Python for example, calculate the latitude and longitude of the subpoint,
latsub = np.radians(dec)
lonsub = np.radians(RA - GST) # where these have already been converted to degrees

Pick a bunch of points to make a circle
theta = np.linspace(0, 2*np.pi, 10001)

Draw a unit circle around Earth's equator, assuming the subpoint is at the north pole (dec = +90 degrees).
x, y, z = [f(theta) for f in (np.cos, np.sin, np.zeros_like)]

Tilt it by 90 - dec
tilt = np.pi/2 - np.radians(dec)
x, z = [x * f(tilt) for f in (np.cos, np.sin)]

calculate the latitude and longitude of the tilted circle by shifting the longitude (subtracting the longitude of the subpoint)
loncirc = np.arctan2(y, x) - lonsub
latcirc = np.arctan2(z, np.sqrt(x**2 + y**2))

Put it all together and convert units properly, and you get something like this. I did it for the Sun but you can do this for any object for which you have the R.A. and Dec.

import numpy as np
import matplotlib.pyplot as plt

# for right now, from Google:
RA = (3 + 29/60 + 38/3600) * 360 / 24 # deg for the Sun
dec = +19. # deg for the Sun
GST = (16 + 15/60) * 360 / 24 # deg

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

latsub = np.radians(dec)
lonsub = np.radians(RA - GST) # where these have already been converted to degrees

theta = np.linspace(0, twopi, 10001)
x, y, z = [f(theta) for f in (np.cos, np.sin, np.zeros_like)]

tilt = halfpi - np.radians(dec)
x, z = [x * f(tilt) for f in (np.cos, np.sin)]

loncirc = np.arctan2(y, x) - lonsub
loncirc = np.mod(loncirc + pi, twopi) - pi
latcirc = np.arctan2(z, np.sqrt(x**2 + y**2))

dloncirc = np.abs(loncirc[1:] - loncirc[:-1])
jump = np.argmax(dloncirc)

latcirc, loncirc = [thing[:-1] for thing in (latcirc, loncirc)]
loncirc[jump] = np.nan
latcirc[jump] = np.nan

locd, lacd = [np.degrees(thing) for thing in (loncirc, latcirc)]
losd, lasd = [np.degrees(thing) for thing in (lonsub, latsub)]
losd = np.mod(losd + 180, 360) - 180 # put it within the maps -180 to +180 range

plt.plot(locd, lacd)
plt.xlim(-182, 182)
plt.ylim(-90, 90)
x, y = 121.5, 25 # Taipei
plt.plot([x], [y], 'or')
plt.text(x, y, ' taipei')
plt.plot([-losd], [lasd], 'oy')
plt.text(-losd, lasd, ' sun')

plt.title('Sunlit Earth 08:40 local time TPE, sunrise was 5:09')
plt.gca().set_aspect('equal')
plt.show()


Extracting kinematics from stellar spectra. Why are template stellar spectra necessary? I am trying to understand how to extract stellar velocity and velocity dispersion maps from spectral cubes, but with no success so far. To be more precise, I do not understand why people use stellar spectra templates to do so.

For gas kinematics (e.g., Ha emission), the usual approach is to fit a Gaussian function to each spectrum. Then, the means and sigmas correspond to the line-of-sight velocities and velocity dispersions.

For stellar kinematics, most people use pPXF. To my understanding, in the simplest case, pPFX fits a parametric function (e.g., Gaussian) convolved with a template stellar spectrum to the data. Why is that? I tried to read the pPFX papers but there is no explanation for this. It assumes that the use of a template stellar spectrum is obvious and there is no justification for this. Actually the majority of papers that use pPFX assume that the use of a template is a de facto thing and provide no explanation.

What is the rationale behind using a template stellar spectrum?
Why can't we just use an inverted Gaussian and fit that to the absorption line of each spectrum?

After some more reading, some papers imply (but do not explicitly say it) that the absorption lines of stellar spectra have complex shapes that can't be modeled with a simple Gaussian. But that is also the case for emission lines! Emission lines can be asymmetric due to multiple kinematic components or beam smearing. Yet, nobody uses templates when extracting kinematic maps from them. I do understand that fitting a single Gaussian to an asymmetric emission line can result in loss of information. But why is this acceptable for emission lines and not acceptable for absorption lines?

(Optically thin) Emission lines in nebulae are generally unresolved and have simple shapes that might be approximated by Gaussians. Stellar spectra are complicated - there are blends, there is rotational broadening, the lines are not Gaussian, there are molecular bands, there may be two stars!

By cross-correlating (or fitting) a template you are effectively modelling your spectrum with what you already think is the best representation of the spectrum.

If emission lines are not unresolved and show velocity structure, as you say, then fitting a simple Gaussian will give an incorrect result. A more complex model would be required, but as far as I know there are no template spectra that can help you out here. The situation might be thought analogous to a stellar spectrum with rotational broadening, but there the template can simply be convolved with the appropriate rotational broadening kernel.


What would the night sky look like from a planet at the centre of a ring galaxy? If there were an observer on a planet in the centre of a ring galaxy (omitting the galactic core) what could they expect to see in their night sky? I am currently under the impression that their night sky would be divided by a dense band of stars stretching across the celestial sphere, similar to the milky band of our sky - though to a greater extent. But is there any reason this wouldn't be the case?

It would perhaps be less impressive than you imagine, and not much different from how any spiral galaxy would look from near the centre.

The many stars near the centre would mean that there were lots of foreground stars and dust near the centre would obscure parts of the ring. The stars of the bulge would mean the whole sky was filled with star fields.

The ring itself would be distant and so quite dim, it would look like a "milky way", but without a central bulge. So it might look something like the milky way in Auriga. Nice, but not especially dramatic.


Convert dates to julian day number in astropy? I'm using astropy and I need to convert dates like this:
2012.06.22_17:56:15_TAI
into julian day number. I tried:

from astropy.time import Time
hmitime = hminew['MidTime']
i=0
print(hmitime[i])
hmi0=Time(hmitime[i],scale='tai')
print(hmi0.jd)

where hmitime it's just the matrix with my dates. But python shows me this:

ValueError: Input values did not match any of the formats where the format keyword is optional [u'datetime', u'iso', u'isot', u'yday', u'fits', u'byear_str', u'jyear_str']

Does anyone know what is wrong? How can I convert my dates??
Thanks

The accepted string formats for date-hms are:

Format  Class    Example argument

fits    TimeFITS    ‘2000-01-01T00:00:00.000’
iso     TimeISO     ‘2000-01-01 00:00:00.000’
isot    TimeISOT    ‘2000-01-01T00:00:00.000’

https://docs.astropy.org/en/stable/time/index.html

You can either change one of these, or you can define your own format by deriving a class for the format from the TimeFormat class.
https://docs.astropy.org/en/stable/api/astropy.time.TimeFormat.html#astropy.time.TimeFormat


Is there a term for asteroseismology as applied to giant planets? Giant planets such as Jupiter have oscillations which enable analyses using the techniques of asteroseismology, for example Gaulme et al. (2011) detected global modes on Jupiter via radial velocity measurements. Is there a specific term for this technique as applied to giant planets instead of stars?

Is there a specific term for this technique (asteroseismology) as applied to giant planets instead of stars?

No it seems there is not.

Instead, people just use normal literal constructions.

"Seismology of Giant Planets""
"Jovian Seismology" and "giant planet seismology" or (your planet here) seismology in general
"Planetary normal-mode seismology"

A good example of the Fourier imaging technique is SYMPA:

SYMPA, a dedicated instrument for Jovian seismology I. Principle and performance
SYMPA, a dedicated instrument for Jovian seismology
II. Real performance and first results

JOVIAL is next-gen implementation, also from Earth:

JOVIAL; Jovian Oscillations through radial Velocimetry ImAging observations at several Longitudes.
Jupiter’s interior: from gravimetry to seismology

note: after posting I realized that one of the linked papers is the same as in the OP"s question. I'll leave it here for completeness.


Is it still the case that the transit of Mercury can't be measured with sufficient accuracy to determine the Sun's distance? The book Transits -- When planets cross the Sun, Maunder & Moore, page 44 states that

Legentil [was] one of the first to realise
that transits of Mercury could never be observed with
sufficient accuracy to make them really valuable in
measuring the length of the astronomical unit.

Is this still the case? With modern instrumentation, can the transit of Mercury still not be measured with enough precision to derive the distance to the Sun? If not, why not? Has Mercury never been used to determine the distance between the Earth and the Sun?

When transits of Mercury and Venus occur, amateur groups do try to observe those transits so as to measure the astronomical unit. To do so, observations of a transit must be made at widely separated points on the Earth. The separated observers will see slightly different versions of a transit, with the differences eventually yielding an estimate of the astronomical unit.
Observations of transits inevitably yield less accurate estimates than do modern estimates based on radar observations of the distances to other planets and on even more accurate distance measurements based on probes sent to other planets. The radar and communications measurements are extremely accurate, twelve places of accuracy or more. Modern transit-based estimates using accurate clocks and GPS are lucky to get better than four places of accuracy.
Transit-based estimates are less accurate than radar-based and communications-based estimates because of the very small parallax involved in Earth-based observations of a transit. Observations of transits of Mercury yield even less accurate results than do observations of transits of Venus because Mercury is closer to the Sun / further from Earth than is Venus during its transits.


Is a black hole made from a anti-matter star the same as a ordinary black hole? Would a black hole from a star made of anti-matter be the same as a blackhole made of ordinary matter?.

To put a formal answer to this question, a black hole made of anti-matter would be indistinguishable from a matter black hole. Black holes are dubious objects, but it seems likely that it makes no sense to talk about an "anti-matter black hole" as once the matter goes into the black hole, be it matter or anti-matter, it ceases to have properties that define it as matter or anti-matter. The No-Hair Theorem for black holes states that, among a few other properties, a black hole is defined only by it's mass and net charge. The matter state of individual particles which fell into the black hole play no part in that description.

In fact, there is a vigorous debate as to whether or not "information" is conserved for black holes. E.g., if an anti-matter particle fell into a black hole, could you, at a later time, pull out a particle of the same properties or would you get out something of completely randomly new properties, not tied to the original matter. This in part plays into your question because it indicates that there is still a lot of uncertainty regarding the loss of information (including if the particle is anti-matter or matter) as particles fall into a black hole and whether not feeding anti-matter into a black hole actually makes it an anti-matter black hole or just simply a black hole.


How can we explain high redshift numbers? I just finished an introductory astrophysics course$^1$ and I have a lingering question that I can't seem to resolve.

We learned that for the first few hundred million years, the universe was pretty boring and not much interesting$^2$ happened. We also learned, correctly or not, that the first stars started forming somewhere between 500 million and 600  million years after the Big Bang.

We also learned how to use redshift values to calculate age, and we talked about how the highest $z$-number we've discovered is some galaxy (GN-z11) at roughly $z=11$.

Using this calculator, we calcultaed that this galaxy apparently formed about 410 million years after the Big Bang.

So, this galaxy seems to be older than when astrophysicists think star formation happened. How can this possibly be? Clearly one of my assumptions is wrong, so is it:

Stars actually started forming before 500 million years post-Big Bang.
Using the UCLA calculator to calculate age is technically incorrect.
A galaxy doesn't need stars to be considered a galaxy.
Some other assumption I made is wrong which makes this post invalid.

As a follow-up question, what happens if we keep finding galaxies at higher $z$-numbers? At what point do we need to reconsider our theory about what happened in the "early" universe?

$^1$I'm not an astrophysics major, so forgive any blatant falsehoods in this post.

$^2$On a macro-scale, at least.

There is evidence that this assumption is incorrect.

  Stars actually started forming before 500 million years post-Big Bang.

There is some evidence that Population III stars formed as early as 200 million years after the Big Bang. The oldest known star, the Methuselah Star, dates back to 200 million years after the Big Bang. It is a second generation star (judging from its metal content), so that means a generation of very large stars must have lived and died before this one was born.


Venus-Saturn conjunction in Sagittarius on Dec 11, 2019 - How often does this happen? Venus and Saturn will be approximately two finger widths apart in the constellation Sagittarius on December 11, 2019. I was wondering: how often do Venus and Saturn come together in the sky like this? I would assume it wouldn't be that frequent. Is it quite frequent? Is there a formula to predict something like this?

To answer How to calculate conjunctions of 2 planets earlier, I computed all planetary conjunctions from -13200 to 17190 and you can search the results at:

http://search.astro.barrycarter.info/index.pl

Prefilled with the values you want (assuming two fingerwidths is 2.5 degrees):

http://search.astro.barrycarter.info/index.pl?p1=Venus&p2=Saturn&lowyear=-13200&highyear=17190&sep=2.5&sunsep=0&submit=SEARCH

and a direct link to the results (with the "LIMIT 200" removed, so you can see all 26,671 results):

http://ae2bcbd33ae927c8e17c62ff155325f3.astro.db.mysql.94y.info/

Note that there are 26,671 2.5 degree conjunctions in 30,390 years, which works out to about 0.88 conjunctions per year.

The http://search.astro.barrycarter.info/index.pl page also provides information on bulk downloading the data, and you may also want to visit:

https://github.com/barrycarter/bcapps/tree/master/ASTRO/CONJUNCTIONS/

Read the README first, but the file you need is the 26.out.bz2 files, since Venus is the second planet and Saturn is the 6th planet.

As always, I'm not perfect, so please don't rely on my results, but do tell me if you find errors, thank you.


What effect does time dilation have on bodies orbiting close to Black Holes? As far as my (somewhat basic) knowledge of astrophysics goes in general the closer to a star your orbit gets smaller (because you travel less distance) and faster (because you're deeper in the gravity well and need to go quicker to avoid falling in.)

But what happens on bodies orbiting close to black holes and subject to time dilation as far as an external observer is concerned?

This question isn't about interstellar but it's a good point of reference. Take the water planet they drop onto where gravitational time dilation is enough to make 20 or 30 years pass to the external observers while only a few hours pass for them. (I don't know if this is a realistic rate(?) of time dilation.)

What would the apparent orbit of their planet be to the external orbit and what would his orbit appear to be for them? Even though he is further from the star would he complete more orbits in the elapsed time despite being further from the centre of the system? And would they appear to be orbiting "slower" than him from his point of view.

Out of interest, though perhaps a second question is more appropriate, how would the apparent difference in relative orbital speed affect rendezvous manoeuvres if at all?

For example would the standard decelerate and fall into a lower orbit still apply but the trajectory plotted from a static observer look weird as time dilation set in? Or would the manoeuvre have to be different to account for the apparent velocity difference? What would a non-circular orbit look like in this system when observed from a circular orbit?

Also is the velocity required for orbit so close to a black hole high enough that speed based relativity also comes into play?

Trajectories close to a black hole can't be approximated by Newtonian mechanics, General Relativity is needed. The black hole doesn't have a surface, but the Schwarzschild radius is point of no return. However it is not possible to orbit just above the Schwartzchild radius.

If you are less than 1.5 radii from the event horizon, there are no orbits. The orbital velocity closer to a black hole than 1.5 radii would be greater than the speed of light. Inside this radius, the faster you try to go, the less the centrifugal force you experience becomes, and so the faster you would fall towards the blackhole.

In fact you won't find a stable orbit closer than 3 radii from the black hole. If the black hole is spinning (as most younger ones are expected to do) the space around the black hole is dragged along by the spinning black hole. It is possible to gain energy by entering this region of dragged space around a black hole. This all means that a space ship can't remain in the region of ultra-intense gravity for many hours by orbiting, It can't experience (from the point of view of a distant observer) extended gravitation time dilation by orbiting a black hole only a few km above its Schwarzschild radius.

It is possible to approach closer than 1.5 radii. An outside observer would see the spaceship approach the black-hole in a spiral at very high speed, but not cross the event horizon. The orbit would not appear to be elliptical and it would appear time dilated. Outside of 1.5 radii the dilation would be less.

Trajectories close to the black hole would have to be calculated using GR. Newtonian mechanics would not be a close approximation. The calculation in GR would be computationally intensive, since the GR equations can't be integrated exactly, even for a two body problem, so a numerical technique would have to be used.

Provided you stay rather further from the black hole, then the usual Newtonian approximation is valid. There's no certain boundary at which Newtonian mechanics is a good approximation. It depends on the accuracy needed.

Remember also that the region close to a black hole is a violent region of space. It would be dangerous for a ship to be so close to a black hole due to the intense radiation produced by infalling matter, and tidal forces.


How long will it be until the nearby groups of galaxies are receding at speeds faster than light? And subsequently how long will it be until we can never see them again?

This will take about 2 trillion years.

Given everything we know now, it's not too hard to predict the eventual fate of the universe. Of course there may be some changes as our knowledge advances, but I think the general course of events will occur as we expect them to. Based on our current knowledge, if you run through the math, you find that after about 2 trillion years from now, the expansion of the universe will have made it so that all by the closest galaxies will no longer be visible to us. You do ask two separate questions, but they're more or less the same question. They might not occur precisely at the same time, but predictions of the future aren't precise enough to distinguish such a small time difference.

By the way, you can read more about the future of the universe on Wikipedia, and specifically this section as it relates to your question.


How can I convert Bayer/Flamsteed designations to Hipparchos numbers? In a text file I have a table of star data with Bayer/Flamsteed designations. I need to convert these into Hipparchos numbers. How can I automatically do this?

As the names and numbers are essentially arbitrary, the Hipparcos designation cannot be calculated from the Bayer or Flamsteed designation.  Instead you have to look up the star in a database.

There is a HD-DM-GC-HR-HIP-Bayer-Flamsteed Cross Index on line.  (Direct link)You can get a computer to do the look up for you. Or if you only have a small number of stars, you can use SIMBAD and do the look-up by hand.


Why can't our Sun be a binary with Jupiter as a T or Y dwarf? I just learned about Brown Dwarfs, they are "failed" stars, they narrowly missed the stellar mass mark. I learned that Y Dwarfs have temperature as low as 80 Fahrenheit (The first one found by WISE observatory), why aren't those not called a Jupiter-like planets without a host star? Why do we call them a star and designate a separate category for them (T or Y)?

Ok suppose T and Y dwarfs are okay, let's leave them alone. But why can't Jupiter be a Y Dwarf who is in the binary relationship with the Sun? Sun-Jupiter's Barycenter is just outside the Sun, we could call them in orbit with each other correct?

But why can't Jupiter be a Y Dwarf who is in the binary relationship with the Sun?

There are two reasons: One is that Jupiter is too small to have ever undergone fusion of any sort. To qualify as a brown dwarf, an object needs to be large enough to have undergone deuterium fusion in its core. This requires a mass of at least 13 Jupiter masses. The other is that Jupiter formed by planetary formation mechanisms rather than by gravitational collapse.

There is debate on what to call objects of less than 13 Jupiter masses that formed by gravitational collapse. Are these sub brown dwarfs or free-floating gas giants, and does it even matter? There is also debate on what to call objects above 13 Jupiter masses that formed by planetary formation mechanisms as opposed to by gravitational collapse. Are these brown dwarfs or hyper planets, and once again, does it even matter?

In any case, Jupiter is not a brown dwarf.


Does Jupiter lose some atmosphere at all? I've hear of something like this: "All planets lose their atmospheres eventually and they become a barren rock if given enough time."

There are three types of atmospheric loss ways:

Jeans Escape: Temperature and escape velocity factors determine the gasses and the amount that's lost. Jupiter cannot lose atmosphere via Jeans escape because Jupiter has too high escape velocity and too high of a temperature.
Charge exchange: Solar radiation creates electrons and positively charged ions in upper atmospheres by tearing electrons off atoms or molecules. Subsequently, charge attraction and repulsion in collisions accelerates ions. Jupiter most likely cannot lose its atmosphere via charge exchange because its magnetic field is strong and extensive while no(little?) material escapes via polar wind.
Vertical atmospheric escape/impact erosion: Energetic objects that strike a planet erodes its atmosphere by creating a plume of heated gas. Jupiter does not lose its atmosphere via impact erosion because it has very high gravity and escape velocity. This pulls particles ejected by impact back into its atmosphere.

Since Jupiter doesn't lose atmosphere in the three types of atmospheric loss, how does it lose its atmosphere?

The phrase you hear is correct! Atmosphere will always be lost, no matter if it is a star or planet. Stars lose their mass via solar wind. Planets lose theirs via the three types of atmospheric loss you mentioned. Frozen planets lost their atmosphere thru sublimation and other things such as ionization. So, yes Jupiter does indeed lose its atmosphere.

You compared the three types of atmospheric loss and how they relate to Jupiter. This is where you assumptions turn out to be wrong. The thermal loss aka Jean's escape affects Jupiter even though it may seem to be not affected in the chart. Periodical or random heat generated by things could excite particles enough to overcome the escape velocity.

The solar wind, even with Jupiter's extensive magnetic field, will reach the atmosphere and neutralize atoms, which in turn will then not be effected by the magnetic field and drift off.

Jupiter is also prone to bombardment from asteroids and comets and other solar system junk. These can range from large sizes such as Comet Shoemaker-Levy 9, to minute small pieces that are barely larger than a pea. These impacts, though they add a considerable amount of stuff into Jupiter, eject mass into space, even with such strong gravitational pull via vertical atmospheric escape, creating a hot plume of material that rises and escapes from a planet. This effect nullifies most material Jupiter gains from the collisions.

Then, will all things added together, we see that Jupiter is gaining mass only via collection of objects, which in turn removes mass from Jupiter via vertical atmospheric escape. This means only a small portion of all mass collected by Jupiter this way is a gain. Now, subtract off thermal and charge exchange escape from the gain, and you will see that Jupiter is not really gaining any material, but rather losing it. Over time, if enough is given, Jupiter will only be a barren core.


R.A. and DEC. for Constellation area vertices I'm developing a planetarium software and I need the R.A. and DEC that represents every constellation area in the sky. I think I only need R.A. and DEC for each vertex.

Where can I find it?

I'm searching on Internet and I can't find it. Maybe because I'm not doing the right search.

James Kilfiger has suggested me to expand my comment as answer, and here it is:

Searching I have found the data in the international astronomical union website. In http://iau.org/public/themes/constellations you can find a TXT for each constellation with vertex's R.A. and DEC.

At the end of that page there is an explanation about the fields on that text files. Their coordinates are J2000.


What are "p" and "r" wrt the moon? Still learning, thinking - and failing - about the moon.
timeanddate.com has this cute and dynamic moon sketch at e.g. https://www.timeanddate.com/moon/uk/london

They generate this based on three parameters that they call i, p and r. The above image is the result of invoking this: https://www.timeanddate.com/scripts/moon.php?i=0.028&p=3.388&r=0.035.
I know that i stands for i llumination in the range 0-1.
However, what are p and r and in what unit? What are the likely correct terms? r appears to be some sort of "rotation angle" of the moon while p is the same for the "mask".

Yes, the are angles in radians. With "i=0.5" to make things clearer:

$p=0$ is the right half illuminated
$p=\pi/2=1.57$ is top half illuminated  (rotated anticlockwise by 90 degrees)
$p=\pi=3.14$ is left half illuminated  (rotated 180 degrees)

and so on.
The r parameter is the rotation of the moon, so $r=0$ is "lunar north up" $r=pi/2$ is rotated anticlockwise by 90 degrees.


Comparing Light Collection Between Pupil and Telescope? The amount of light collected for a period of time is proportional to the area of the collecting aperture. We use our eyes -pupils have diameters of approximately 5 mm-, binoculars and telescopes. How much more light can we collect with a 50mm pair of binoculars compared with our eyes? And with an 8-inch telescope? If we could have access to a 10-meter class professional telescope, how large would the gain be?

50^2/5^2= 100 times more light

8in = 203.2 mm

203.2^2/5^2= 1652 times more light

10000^2/5^2= 4000000 times more light

Did I solve this problem correctly?

You have the idea right! The amount of light collected is indeed proportional to the area of the collecting surface.

Now, if you want to take it a bit further, try to think about the following:

Binoculars have one lens for each eye, does this mean you get twice the amount of light in the end?
Reflecting telescopes almost always have a secondary mirror which results in part of the light being masked and not collected by the primary mirror. Now imagine you have an 8-inch telescope with a 2-inch secondary mirror, can you calculate again the light collection ratio compared to the naked eye?


Why do things float in space, though the gravity of our star is always present? In the solar system, things float around. We are so certain that the gravity of our sun exists. Still, why does that gravity not influence satellites and other objects there?

Our solar system is also orbiting the center of our galaxy, and the gravity of our black hole, but still many objects float around. I wonder why.

It is not true that "objects float around" in the solar system.

Perhaps you have seen video from the space station, and you can see things floating. This is not because there is no gravity, but because everything in the space station going at the same speed in the same direction. This makes it look as if things are floating. In fact the space station and everything in it, is moving at about 7km per second around the Earth.

The Earth and everything on it is is also moving at 30 km/s around the sun. The satellites around the Earth are also affected by the sun's gravity, but because they are affect the same as the Earth, they are also moving at about 30km/s around the sun. If two things are moving at the same speed and in the same direction, it will look as if they are floating. In fact they are "whizzing".

The Sun, Earth and satellites are also whizzing around in the gravity of the galaxy.  The black hole is only a very small part of this; most of the gravity of the galaxy is in the dark matter that we can't see.  But we don't feel this pull because the Sun, Earth, the satellites and you are all being pulled at the same time.


Could a magnetars magnetic field have saved Mars atmosphere? Just wondering: if a magnetar was close enough to Mars would it have saved its atmosphere by protecting it from solar radiation? I was wondering this because Earth protects us via a magnetic field; the magnetar would be providing Mars one. This is if it did not destroy the planet first as suggested here.

Edit
I meant if the magnetar was safe to live by.

How would you get a magnetar close to Mars, within the solar system?

OK, let's leave that to one side, but the Earth's magnetic field is of  order $10^{-4}$ Tesla at strongest. As the dipole field of a magnetar diminishes as $r^{-3}$ and could be as high as $10^{11}$ Teslas at the surface of the magnetar, at a radius of 10 km, then the magnetar woud need to be at a distance of $10^{6}$ km to surround Mars with a field of $10^{-4}$ Tesla.

The problem is that the radiation coming from the magnetar in the form of charged particles accelerated from the magnetic poles would completely dwarf the charged particle radiation from the Sun at Mars.

So no, a nearby Magnetar would not save Mars' atmosphere - quite the reverse.

EDIT: A BOTE calculations: If we look at the Crab pulsar - not quite a magnetar, but close - it is calculated that 10% of the rotational KE being lost ends up in synchrotron raiation from charged particles. That's about $10^{30}$ Watts. So there is at least $\sim 10^4$ solar luminosities, just in the kinetic energy of relativistic particles.

Magnetars are also very strong and variable sources of hard X-ray and Gamma ray radiation. Typical magnetar bursts/flares can release $10^{7}-10^{9}$ solar luminosities of hard X-rays and Gamma rays over periods of 1-40 s. That should fry any atmosphere (or anything else) that was only $10^{6}$ km away!


What causes the strong winds on Uranus? I was just looking up something about Uranus on Google, and read a line I had read many times before - "The strong winds of Uranus...". So, short and easy - "What causes the strong winds on Uranus?".

As said, winds on Uranus are quite strong blowing upto 560 miles/hr (900 kmph). But, the question is what forces drives these winds?
Researchers has been digging for quite a while. They investigated the gravity fields of Uranus using data gathered by NASA's Voyager 2 spacecraft and ground-based telescopes. The strength of a planet's gravity field depends on its amount of mass, and this strength can vary over the surface of a planet depending on the amount of mass lying under it.  By analyzing the gravity fields of these worlds, the investigators could deduce how their atmospheres circulated. From the readings, they deduced that the winds blow in relatively thin weather layers no more than 600 miles (1,000 kilometers) but comparing to size of Uranus, the layers are proportionally very small (0.15 % of the mass of Uranus). So, the researchers propose that the winds are mostly driven by atmospheric and surface conditions (possibly moisture evaporating and condensing in atmosphere causing the winds to swirl).
Some suggest that since the outer planets are gaseous, there is virtually no surface roughness to act as a drag on winds like on Earth and hence the atmospheres is more fluid like. Moreover, Uranus is very far away from the sun and thus there is less solar energy to impact the turbulence in the atmosphere.
Some also suggests internal heat to be the reason. Uranus only puts out about 6 percent as much heat as it receives from the Sun. So, internal heat source might be involved on some level.
For more information, read this excellent paper: Atmospheric confinement of jet streams on Uranus and Neptune
Additional source:

https://www.space.com/21157-uranus-neptune-winds-revealed.html
https://phys.org/news/2013-05-uranus-neptune-confined-thin-atmosphere.html
https://www.csmonitor.com/Science/2013/0517/Why-do-planets-farthest-from-sun-have-highest-winds-Team-closes-in-on-answer
https://arstechnica.com/science/2013/05/powerful-winds-on-giant-planets-confined-to-a-thin-atmospheric-layer/


Trans-Neptunian Dwarf Planet near Heliopause Is there any reason to think a Trans-Neptunian dwarf planet, the size of Pluto, could not exist in a Keplerian orbit 120AU from the Sun?

If such an object existed, and came within 0.1AU of the heliopause, is there anything about the environment near the heliopause that would prevent a stable orbit.

Heliopause is believed to be at a distance of about 123 AU away from the Sun. Now, if you look at the list of farthest body in the solar system, you will see two bodies located approximately at that distance: 2018 VG18 and predicted FarFarOut (124.12 AU and ~140 AU from the sun as of June 2020 respectively). They are considered the farthest TNOs in the solar system.
For comparison, Voyager 1 and Voyager 2 are the two space probes which left the solar system is now 149.36 AU and 124 AU from the sun as of June 2020. Also you can see some of the near-parabolic comets originating from that distance.


A good telescope for the viewing of Nebulae, Stars and Planets So, I am a 15 year old interested in astronomy. My father used to have a reflector telescope, but was never so serious about it that he could recommend me a good telescope. I am looking to buy a telescope up to £400 (possibly £500) and would like one that is mainly great for looking at nebulae. Stars and Planets would be great also, but are not necessary. Light Pollution isn't too bad. I have been recommended a Newtonian Reflector, but am not too sure if this would be suitable for my conditions. Also, does anyone recommend a good, British website where I could buy such a scope from? Finally, is there much point in me buying a computerized scope, will this allow me to align the scope better, or is it just going to be a waste of money.

This is a very common question, yet very hard to answer if you prefer a clear, concise, uncontroversial answer that applies to all situations. So I'm not going to do that.

Instead, I'm going to describe your main options, and let you choose. Be aware that you'll make the choice while still not knowing much about optics. So, in a sense, it will be just the beginning of a learning journey, and you'll have to keep learning and trying things out. Hope you like that sort of thing, because it's kind of mandatory in this hobby.

Also, there are two kinds of people in this field: those who enjoy using the instruments, and those who enjoy making the instruments. I'm 25% of the former, 75% of the latter kind (I make telescopes and mirrors, and I do some optics design). So that's the perspective you'll get.

First off, a few words about the objects you'll observe. There are two kinds of objects:

HIGH RESOLUTION, HIGH BRIGHTNESS

These are the planets, the Moon, many double stars, and the like. These are typically pretty bright, so the amount of light you capture is not essential. But these objects show high-resolution detail, so the resolving power of your scope is crucial.

You get more resolving power if the aperture (the diameter) of the scope is bigger. But there's a problem - seeing (the inverse of turbulence).

http://en.wikipedia.org/wiki/Astronomical_seeing

When air is turbulent, it will reduce the resolving power of your instrument. If your scope is 200 mm (8") or bigger, this will happen A LOT. If your instrument is 100 mm (4") or smaller, it's not so noticeable because you don't have that much resolving power to begin with.

There are ways to forecast seeing, because it depends on weather:

http://cleardarksky.com/csk/

So, for high resolution targets, the point of having a large instrument (like a huge dobsonian) is because once in a while seeing is so good (turbulence is so low) that you can actually use all its huge power - but most of the time you'll be using it at much less than its theoretical power, as if it's a 4" instrument or whatever the seeing is at the moment.

These objects are not affected by light pollution. You could observe the Moon from a brightly illumined parking lot in the middle of the city. You don't need to be in deep darkness while observing them - in fact, it's beneficial if you're not adapted to very deep darkness.

LOW BRIGHTNESS

These are most of the nebulae, star clusters, and galaxies. These are distant, faint objects. Their light is very faint, so it's essential you capture as much of it as possible. Again, bigger aperture is better.

These are the kind of objects that huge dobsonians (half-meter aperture, or more) are made for. In a very large scope, there's galaxies everywhere.

But again, there's a catch. These objects are affected by light pollution.

http://www.jshine.net/astronomy/dark_sky/

You need to take your scope far and away from any city, town and industrial complex (at least a 1 hour trip at highway speed). This is pretty hard to do in Europe; in the US, population density is low and this is somewhat easier. Anyway, just stay away from city lights and the view will improve.

A few of these objects will be visible from the city: the Andromeda galaxy, the M13 cluster, the Orion nebula, etc. But many of them look much, much better away from light polution. In any case, you should be able to observe the whole Messier catalogue (over 100 objects) from the city - but out in the boondocks these things look better.

http://en.wikipedia.org/wiki/List_of_Messier_objects

It's important that your eyes are adapted to very deep darkness when observing these things, so stay away from any source of light. Even the screen of your phone is far too bright. Street lights 500 meters away are too bright. And turn off that flashlight and keep it off. One second of exposure to a bright light destroys a dark adaptation that took 20 minutes to build.

Since these objects are so faint, the eye doesn't see a whole lot of high-resolution detail, so they are not affected by seeing (as far as the human eye is concerned).

Okay, now here are the main types of instruments that you're most likely to encounter:

NEWTONIAN REFLECTORS IN GENERAL (DOBSONIANS IN PARTICULAR)

Many people will say "just get an 8 inch dob as your first scope", and that's usually correct. This design gives the biggest aperture per price. Biggest bang for the buck. It scales up easily to very, very large sizes. It's the ideal instrument for observing the "faint fuzzies" (nebulae, galaxies). It's great for learning, because you can easily observe its guts. When seeing is good, a large dob will open up the sky for you - you can crank up the magnification a lot (depending on aperture).

On the flip side, due to the low price, you will see many dobs out there that are shoddily made.

You need to remember two things with a dob: collimation, and thermal management.

Collimation means keeping all the optics aligned. The mirrors will move a little, due to the way a dob is built, and they'll lose collimation. So don't forget to re-adjust it once in a while. In time, you'll get so good at it, you'll give it a two-minute quick adjustment every time you observe, and it'll be good to go; that's what I do.

This is why collimation is important:

http://legault.perso.sfr.fr/collim.html

This is a good introduction to newtonian collimation (which is a vast topic, you could write a whole book on it):

http://www.cloudynights.com/documents/primer.pdf

Thermal management means keeping the scope at the same temperature as the ambient air. If you don't do this, internal convection will destroy the image. This is why:

http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/

http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/

Small dobs (up to 6...8"), just take it outside at least 1 hour before you observe - or better yet, 2 hours. Let it "breathe". Dobs 10" or larger, you MUST use a fan on the primary mirror to force cooling.

REFRACTORS

These instruments are very convenient and very low maintenance. Take it out of the box, and it's good to go. It's collimated from factory, and it doesn't care about thermal issues. A refractor makes a great "travelscope", and a great grab-and-go scope.

On the flip side, they are the most expensive, in terms of how much aperture you get for the money you spend. And they are typically made in small apertures; most commercial refractors are not bigger than 6" or so; those that are bigger are extremely expensive. The small aperture limits the maximum performance of these instruments.

There is this myth out there that refractors are somehow magically better for planets (and other high-resolution targets), whereas dobs, even very large ones, are not so great. This is false; load up the designs in Zemax, do a basic performance analysis, and you'll see that aperture is the dominant factor, with everything else being secondary. But there are several reasons why this myth is around, and it's instructive to see what these reasons are:

refractors perform at 100% of their potential no matter what, right out of the box. Dobs require maintenance and fiddling to perform at 100%.
many dob owners simply don't do the required maintenance - they don't collimate their scope, and don't give it time to reach thermal equilibrium. Then of course it will perform miserably on planets. There are 12" dobs out there that perform worse than 4" refractors because of how out of collimation they are and how hot the mirror is.
many refractors out there are quite well made (they ought to be, given the higher price you're paying), whereas many dobs are cheaply made in a sweatshop. So you're not comparing apples to apples.

Bottom line: don't buy a refractor because you "want to watch planets". But do buy a refractor if simplicity of use and the low maintenance are very important to you (and you don't care much about price).

A decent refractor on an excellent mount can do pretty good astrophotography. More on that below.

CATADIOPTRICS (CASSEGRAIN, MAKSUTOV, DALL-KIRKHAM, RITCHIE-CHRETIEN)

Broadly speaking, these are for astrophotography (not always, but usually). But can you do visual astronomy with them? Sure. It's just that it's easier to optimize them for photo, with this design.

Usually, they come installed on a motorized mount. All but the very cheapest ones have a computer on-board; you just punch the name of the object into the remote control, and the scope will turn around automatically to face the object. Sounds cool, right?

Well, I am going to say something a little controversial here: this is not a good place to start.

First off, astrophotography has an extremely steep learning curve. It's very easy to take a little blurry photo of Jupiter that will impress nobody. It's very hard to take a great, high resolution photo like these:

http://www.acquerra.com.au/astro/

The skills and knowledge required to take that kind of photos are more easily acquired by doing purely visual astronomy, on a dob or whatever, for a couple years or more. Also, the equipment required for that sort of stunt costs many thousands of dollars - your cheap little Cassegrain cannot do it.

So why buy a small Cass? Well, if you have an overwhelming interest in astrophoto, and are prepared to deal with the difficult learning, then go ahead. But realize that the first images are not going to be very impressive, and the motorized mount will not teach you some valuable skills that you would acquire more easily pushing a dob around and keeping your eye plugged into the ocular for a while. Also, money spent on the motorized mount is money NOT spent on aperture, and the aperture is so important for overall performance.

For astrophotography, the mount is of huge importance; the scope per se is relatively less important (I'm generalizing). A so-so scope on a great mount is much better than the other way around. But a great GEM mount can be very expensive.

Well, that's it. Do some more research on your own, then make a decision. Don't forget that learning only begins with the moment you acquire the scope.

Above all, have fun, and keep looking up.

Clear skies to you!