Roche Zone of a black hole vs Roche Zone of Neutron Star Suppose a very unusually small back hole with the mass of about, lets say, 1.5 solar masses(hawking radiation should destroy a lot of mass let's assume). It thus would have a Roche limit smaller than larger black holes, and the Roche limit should be outside of the black hole instead of inside it, like some supermassive black holes.

Then, let's take a super massive neutron star, perhaps even with quark matter at its center. Even though this may not exist, let's say that is 3 solar masses by taking matter from a nearby red giant rather slowly. This would mean that it thus has a larger Roche limit.

Then, let's fit the two together in a binary system. Move the black hole inwards so that it is in the Roche limit of the neutron star, but its Roche limit is not affecting the neutron star.

My question is that, since a black hole can't physically be ripped apart by gravity due to it being a hole in space time itself, can it's structure be altered in any way by the gravitational pull of the neutron star? Like from a sphere to an oval or something like that? Can it separate into two smaller ones due to the gravity of the neutron star? In the system, who would win the gravity war?

Now assume that the black hole has angular momentum and change, giving it a magnetic field. Then make the neutron star a magnetar of the strongest kind, and place them in the same binary position as I described above. Would their magnetic fields have an effect on any of the bodies that are generating them? Would their magnetic fields have any effect on the other object's magnetic field?

The Roche limit is defined in classical physics, dealing with materials that behave classically. That said, of course people have tried to estimate the counterpart in relativistic conditions. It turns out that it behaves roughly like the classical case when considering a liquid body orbiting a black hole. For the neutron star case and a more massive stellar mass black hole the limit is not too far away from the ISCO.

Would a black hole be affected by tidal forces? A tidal force is after all just the effect of the radially varying curvature of spacetime, and a black hole embedded in such a curved spacetime does show distortion. Not only that, but one can apparently treat the black hole as a fluid sphere that can even dissipate! (paper, presentation2) As the plot in the second presentation shows, yes, the horizon can get distorted by a nearby massive orbiting object.

However, the laws of black hole thermodynamics do not allow the horizon area to go down. Splitting a mass 2M hole into two holes of mass M would change the area from $4 \times (4\pi G^2 M^2/c^4)$ to $2\times (4\pi G^2 M^2/c^4)$. So that is not going to happen (unless there was a process that could release a black hole's worth of entropy in the outside universe - a very tall order).

In the end, what would happen is that the deformations would dissipate orbital energy (by heating the neutron star and gravitational radiation) until the objects spiralled together, the neutron star would cross the relativistic Roche limit and shortly after the remnants would plunge inside the ISCO and be absorbed.

The magnetic field case would complicate things in various ways, but remember that even magnetar magnetic fields only have the density of $10^4$ times lead - compared to the neutron star density this is nothing. There would presumably be a lot of electromagnetic dissipation, jets, and complex torques, but I suspect the endgame is the same.


Questions about Planck’s law Two questions:

I read online that all objects emit the majority of their thermal radiation in the radio, and that this emission increases with temperature. Does this mean that all glowing bodies emit the majority of their radiation in the radio, regardless of their temperature?
Wien’s law is used to find the peak wavelength of a black body. Does this mean that there is zero emission at any wavelength shorter than this peak wavelength?

Thank you in advance for your help.

It would be good to specify what "majority of radiation" means. If it means "radiative energy", then this premise is surely wrong. To find the energy emitted per wavelength, take the Planck function $B(T, \lambda)$ at a certain temperature $T$ and multiply with the wavelength $\lambda$. You will find that the short-wavelength radiation always dominates for a black-body.
Wien's law gives you the peak of the Planck function. Radiation at all other wavelengths exists.

Note that objects predominantly loose most of their energy, if photons and matter are in equilibrium. If this equilibrium is disturbed, or particle accelerations play a role (like in the presence of magnetic fields), a whole zoo of non-thermal processes can also lead to important radiative energy loss.
Non-thermal processes in space can be radio emission from gyrating particles, line radiation, Masers, Bremsstrahlung, Compton radiation... Those processes then do not show a Planck spectrum, but other characteristic spectra.

Edit: I've added a plot of the Planck function at two different temperatures. In a linear plot (on the left) it might look as if there's a cut-off, but in a logarithmic plot (on the right), one sees what mathematical analysis already betrays: The function is defined everywhere. It just becomes really small on the high-energy end, as a commentator pointed out correctly.


Why do some neutron stars become pulsars? I suspect this may be an open problem in astronomy, but I have to ask anyway.

There doesn't seem to be any large-scale difference in the formation of pulsars from the formation of "normal" neutron stars. Yet pulsars end up emitting electromagnetic radiation, while others - "radio quiet" neutron stars - do not.

Why do are some neutron stars formed as pulsars, while others are not?

This is partly based on What causes a star to become a pulsar?, but the answer does not cover this part.

My understanding is that it would be quite usual for the majority of neutron stars to be pulsars shortly after they formed.

The requirements are that they should have strong magnetic fields and rapid rotation (faster than a rotation period of a few seconds). The latter seems an almost inevitable consequence of conservation of angular momentum and the former must surely be assisted (if not fully explained) by conservation of magnetic flux as the core of a massive star collapses (see the discussion in Reisenegger 2007).

But to be observed as a pulsar, the spin and magnetic field axes have to be oriented in a specific way, so that Doppler-boosted and beamed radiation sweeps over the Earth. This requirement means that we can observe only a fraction of the pulsars out there.

In fact there are a category of "radio quiet pulsars" - these are detected by their high energy radiation pulsations but not seen as radio pulsars. The idea here is that the beaming is less extreme for the high energy radiation, allowing it to be seen.

The pulsar phenomenon is also transient. Once the rotational kinetic energy of the pulsar has been extracted and it spins down beyond the "pulsar death line", or if magnetic field decay occurs rapidly, then the pulsar should switch off. Most of the billion (or so) neutron stars in our Galaxy are no longer pulsars. The slope of the death line is such (e.g. see Arons 1999) that even if the pulsar is born with a relatively weak field ($10^{6}$ T !) it should still be born with a rotation period short enough ($<1$ s) to produce the pulsar phenomenon initially.

You might find the discussion/simulations in this paper interesting.


What is the probability that there is life on other planets? From a mathematical point of view,  what is the probability that

1) There is organic life on other planets

2) There are advanced species like us or even greater than us on other planets?

Is there a formula or equation or a way we can calculate this probability?

The probabilites are unknown at the moment (March 2014), since there is only one known planet (Earth) harboring life. This doesn't allow any meaningful probability estimates for the occurence of life, based on empirical data.

The overall formation of life is too complex to allow simulations based on current technology.
Although some intermediate steps can be simulated or performed by experiment, e.g. by the Miller-Urey experiment.

The Drake equation is a simplified attempt to decompose the probability into factors. Some of the factors aren't known yet. But there has been much progress in estimating the frequency of exoplanets, and the probability of planets to be habitable. From this news release one could estimate (order of magnitude), that there is one habitable planet around a Sun-like star per about 1,700 (about 12³) cubic light years in our region of the Milky Way ("... the nearest sun-like star with an Earth-size planet in its habitable zone is probably only 12 light years away...").

We don't know the probability for habitable planets to develop life, nor to be colonized. Probabilities of complex life forms are even more difficult to estimate (yet). More discussion about habitability of exoplanets on Wikipedia.


Difference in stellar abundance numbers What is the correlation between stellar abundance by mass and by number?
Let’s take the Sun and Helium for example. This paper mentions an abundance by mass for helium as Y=0.275 and by number as A=10.99, which I believe is derived by assuming N(He)/N(H)=8.5%.

How do you get this 8.5%? Is there a standard value for N(H)?
Why are abundances  given both by mass and by number? Is there a historic reason?
Is there a relation between the two, that is can you estimate one “type” by knowing the value of the other?

According to Lodders (2003, https://arxiv.org/pdf/1010.2746 ) the relative abundance of helium to hydrogen is $A({\rm He})=10.925$, on a logarithmic scale where the hydrogen number abundance is 12. So this would mean a helium to hydrogen ratio, by number, of $10^{10.925-12}=0.08414$. i.e. 8.4% (your source uses 10.93, not 10.99, hence a very slightly different percentage).
Why have the two systems? The appearance of the spectrum - absorption lines, emission lines etc. depends on their number density. So it is relative numbers of the elements that count. On the other hand when you are doing structural calculations on stars, it is usually the mass fractions that matter.
The relationship between the two?
The relative numbers of helium and hydrogen are easily translated into relative mass fractions of helium ($Y$) and hydrogen ($X$):
$$\frac{Y}{X} = 0.08414 \times 4.0026/[(1-0.08414)\times 1.0078] =0.3649$$,
where $4.0026/1.0078$ is the ratio of atomic masses of helium to hydrogen.
Now there is also a small fraction by mass of heavier elements $Z\simeq 0.014$, where $1 = X + Y +Z$.
If we substitute $X= Y/0.3649$ and $Z=0.014$, then
$$ Y \left(1 + \frac{1}{0.3649}\right) = 1-Z$$
$$ Y = 0.264$$
To sum up:
$$Y = \frac{1-Z}{\left(1 + \frac{1.0078(1 - 10^{A({\rm He})-12})}{4.0026\times 10^{A({\rm He})-12}}\right)} = \frac{1-Z}{\left(0.7482 + 0.2518\times10^{12-A({\rm He})}\right)}$$


Can gravity cause a Big Crunch, if the cosmological principle holds? Say we have a closed universe that follows the cosmological principle. As a result, mass is distributed equally and, if as far as I know, gravity should have no net effect on the movements of galaxies (as in, there are no peculiar velocities).

However, in the Big Crunch, the Universe stops expanding and gravity causes the scale factor to decrease. How is this possible? Shouldn't gravity have no effect on a homogenous and isotropic universe?

On that note, if gravity actually caused a collapse, would gravity cause the comoving coordinates of the galaxies to change?

You are thinking about gravity and how it affects the universe as a whole in the wrong way.

The dynamics of the universe are governed by the Friedmann equations, which can be solved to say how the scale factor of a homogeneous, isotropic universe behaves with time.

If there is enough gravitating matter/energy density in the universe then the scale factor may eventually decrease to zero in a finite time. The simplest case is a closed universe with no cosmological constant and a mass density greater than the critical density (which is not the universe we currently inhabit: our universe appears to be (a) flat and (b) expanding at an accelerating rate).

The interpretation of this is that the space between the galaxies is contracting. It is the reversal of what we currently observe in the expanding universe.

In the same way that you do not interpret the universal expansion of a homogeneous, isotropic universe as some sort of repulsion between galaxies; neither would the contraction be interpreted as some sort of attractive force between galaxies. As you rightly point out, in a strictly Newtonian sense, there would be no net gravitational force on any one galaxy and so its co-moving coordinates would be unchanged during the contraction.


Are we still in the Solar Maximus stage? Will this winter 2013/14 be a good aurora-viewing season? I remember reading in 2011 that 2013/2014 would be the peak of the current solar maximus, but I want to confirm whether that is still ongoing.  Will the upcoming winter season be a good time to see the Northern Lights?

From this data here:

http://www.swpc.noaa.gov/SolarCycle/

http://solarscience.msfc.nasa.gov/SunspotCycle.shtml

We are in a local maximum but the actual values have been lower than predicted for this time.

As for getting an idea for the aurora, I would keep an eye on the sunspot numbers and the space weather.

http://www.swpc.noaa.gov/today.html

http://www.gi.alaska.edu/AuroraForecast


Does the CMB pattern evolve in a human lifetime? I was thinking, as CMB is a snapshot of a slice of the Universe during recombination, how much the cmb is changing with time ?
I mean as we progress in time we look a CMB a bit more faraway each time, could this effect be seen and could we watch the pattern of CMB anisotropies changing with time (on a "human" timescale I mean) ?

In principle yes, in practice no.

As seen in the temperature power spectrum below, the Planck satellite detects power (i.e. "a signal") even on the smallest probed scales, which is a few arcminutes.

Fig. 1 from Planck Collaboration et al. (2016) with my own approximate corresponding angles annotated in red.

The smallest scale shown in the figure is $\ell = 2500$, corresponding to an angle of roughly $4'$. At the redshift of the CMB ($z=1100$), this angle subtends 57 million lightyears (Mlyr) today. When the CMB was emitted, such a region was only 52 thousands lightyears (klyr) across).

In other words, we can see that at least down to regions that were roughly 50 klyr across, the CMB pattern is anisotropic.

As time goes, the CMB we see will come from an increasingly larger distance. Hence, when this distance has increased by roughly 50 klyr, the CMB pattern should have evolved. This will take a bit more than 50 thousand years$^\dagger$(kyr), i.e. much longer than a human lifetime.

To be able to see changes within 100 years, we would need an angular resolution of 0.5 arcseconds. But angular resolution is not enough — the power seems to fall off as a power law on these scales, so a higher sensitivity would also be needed.

$^\dagger$A bit more, because space expands somewhat during the time that light travels that distance, so in total roughly 55 kyr.


Moon SHAPE calculator I have seen a lot of information about moon phase calculation, but need an accurate moon shape calculator/algorithm given the observer's position and time. This implies that the illuminated portion may be rotated, and not only have a right-left orientation as traditionally depicted.

There is an accurate depiction of the moon phase and shape in Stellarium (see below) that you might find useful. You can select your observing location and time anywhere on Earth (or another planet if you so wish). Then you can just search for the moon in the search window (if you cant spot it in the sky), and you have a real representation of how the moon looks at that specific time and location. However if you need to repeat this calculation many times then this isn't the best solution.


"How the Universe Works": is it scientifically accurate? The documentary/mini-series How the Universe Works is based on scientific evidence and according to what is known nowadays, or sheer bunk?

Throughout the series, are there any particular topics which are still disputed among the scientific community and that were portrayed in a different light than they should've had been?

I've only watched a few episodes but they always talk in a way that gives you the feeling everything is well-established and accepted but that might not be the case. I figured that an astronomer that watched the series would have remember if this happened or not.

I understand the subjects were researched and there is footage of known scientists in the episodes, but that doesn't mean everyone agrees with them. (By everyone I mean the most part of scientists working on those subjects, and not literally every single one of them.)

To give you an example, on the second episode they state that there are massive black holes at the center of galaxies. From my perspective (i.e. of someone which is not an astronomer) if they say so it's because it's probably true. However, in the real world maybe this is a highly disputed theory and a lot of astronomers disagree with this stance, and they're note telling the audience about it. Why would they do that? I don't know. I don't know if they did, I'm just asking if someone knows whether it happened, or if anyone saw something in the series with which they disagree.

I don't think that the Discovery Channel would produce science fiction and call it a documentary. The episodes are bases on factual information at the time of production.


Early universe's first starlight (indirectly) detected for the first time, could it have been detectable with 1970's technology? The recent  paper in Nature An absorption profile centred at 78 megahertz in the sky-averaged spectrum seems (to me at least) to be an incredibly simple and elegant experiment with far-reaching implications.
While the system has been implemented very carefully, it is essentially a spectrally resolved radiometer at radio frequencies below 200 MHz, measuring the sky-averaged power spectrum. And while many of radiometers in space and on Earth measuring the Cosmic Microwave Background are fairly broad band, measure in the 10's to 100's of GHz range, and intended for very precise absolute intensity measurements of a smooth Plank distribution, the system described here measures between only 50 and 200 MHz, and with a spectral resolution of only 6.1 kHz!
The small set of carefully designed and optimized, antenna(s) and receiver(s) are simple and straightforward; a broadband dipole-type antenna above a large ground plane in a radio-quiet zone, some low noise electronics and what non-radio astronomers might call a very large bandwidth Software-defined Radio or SDRs.
And yet this small, simple receiver, sitting quietly on the ground in Australia's Outback has produced observations are amazing! They've detected direct evidence of the first stars producing the first starlight in our early universe only about 200 million years after the Big Bang!

Abstract:
After stars formed in the early Universe, their ultraviolet light is expected, eventually, to have penetrated the primordial hydrogen gas and altered the excitation state of its 21-centimetre hyperfine line. This alteration would cause the gas to absorb photons from the cosmic microwave background, producing a spectral distortion that should be observable today at radio frequencies of less than 200 megahertz1. Here we report the detection of a flattened absorption profile in the sky-averaged radio spectrum, which is centred at a frequency of 78 megahertz and has a best-fitting full-width at half-maximum of 19 megahertz and an amplitude of 0.5 kelvin. The profile is largely consistent with expectations for the 21-centimetre signal induced by early stars; however, the best-fitting amplitude of the profile is more than a factor of two greater than the largest predictions2. This discrepancy suggests that either the primordial gas was much colder than expected or the background radiation temperature was hotter than expected. Astrophysical phenomena (such as radiation from stars and stellar remnants) are unlikely to account for this discrepancy; of the proposed extensions to the standard model of cosmology and particle physics, only cooling of the gas as a result of interactions between dark matter and baryons seems to explain the observed amplitude3. The low-frequency edge of the observed profile indicates that stars existed and had produced a background of Lyman-α photons by 180 million years after the Big Bang. The high-frequency edge indicates that the gas was heated to above the radiation temperature less than 100 million years later.

I don't see anything about this data that strikes me as unmeasurable with even say 1970's technology if done carefully. The signal is huge, a distinct, broad (about 19 MHz centered at about 87 MHz), several percent (almost 0.6 K) deep dip in an otherwise Rayleigh-Jeans regime blackbody distribution. In 20:20 hindsight, it seems to me this could have been detected a long time ago.
Question: Am I missing something, or is this discovery remarkable not only for it's far reaching implication, but for its simplicity, elegance, and the fact that it could have been discovered quite a long time ago with '60's or '70's technology?

above: Figure 1 from the paper.
below: Figures 2 and then 1 from the paper's Extended data.

I could not find a freely accessible version of this article, so I am mostly guessing here.
From this article https://arxiv.org/pdf/1206.0267.pdf (especially sections 4 and 5), the main challenge seems to be distinguishing between the different components of the signal:

  (1) the 21 cm signal coming from the high redshift Universe,
  (2) galactic and extra-galactic foregrounds,
  (3) ionospheric influences,
  (4) telescope response effects,
  (5) Radio frequency interference and
  (6) thermal noise
  Distinguishing between all these components means understanding each of them equally well and probably requires recent knowledge (in cosmology, radio receptors, etc)

In any case, the idea of using the 21-cm line to probe the early universe seems quite old:
https://academic.oup.com/mnras/article/188/4/791/999622


Does Flat Universe mean even distribution? I've read through all the other related answers to "flatness" questions but I need a bit more clarification.

I understand that a triangle in a 2D universe would not equal 180 degrees in both a closed and open universe. If I am correct, not being flat during the initial conditions would mean exponential expansion or collapse as time went by.

Given the above, could we explain flatness as uniform density of the universe, meaning all of the universe is the same approximate "consistency" on a large scale (homogeneity).

No, being flat and being homogeneous is not equivalent$^\dagger\!\!\!$.

Flatness refers to the geometry, which depends on the total energy density $\rho$; if it is above or below a certain critical threshold $\rho_\mathrm{cr}$, we call the Universe "closed" or "open", respectively, while if $\rho$ is exactly equal to $\rho_\mathrm{cr}$, we call it "flat".

Homogeneity refers, as you say, to matter being evenly distributed (on large scales). If the Universe is not only homogeneous, but also isotropic (i.e. looks the same in all directions), then there are three different possible geometries, namely flat, open, or closed. But a homogeneous universe doesn't have to be isotropic, and both open and closed universes can have evenly distributed matter.

Thus,
$$
\mathrm{homogeneity} \nRightarrow \mathrm{flatness}
$$

Whereas the local geometry depends on the local density$^\ddagger$, the global geometry does not depend on how the matter is distributed. For instance, you could in principle have a universe with no upper limit of structure size. In our Universe, we find observationally that the largest structures have sizes of roughly half a gigalightyear. Above this scale, it is homogeneous (although it could be inhomogeneous on scales larger than the particle horizon). But another universe might have structure on all scales and thus not be homogeneous, but still meet the criterion $\rho = \rho_\mathrm{cr}$ and thus be flat.
You could also, as discussed in this answer about homogeneity and isotropy, imagine a universe originating at a central point away from which the density always decreases (i.e. is isotropic around this point), but which still has $\rho = \rho_\mathrm{cr}$.

Thus,
$$
\mathrm{flatness} \nRightarrow \mathrm{homogeneity}
$$

Note that while your statement about the geometry determining the fate of the Universe would be correct if it contained only matter and radiation, it seems that $\sim70$% of the density making it flat has the annoying propety of accelerating the expansion of the Universe. Thus, whereas a flat, matter-dominated universe would expand asymptotically toward a finite size, our Universe is dominated, it seems, by something dubbed dark energy making it expand exponentially despite being flat.

$^\dagger$For a universe, at least. For a roadkill, it might be equivalent.

$^\ddagger$For instance, space "bends" enough around a massive cluster of galaxies to make gravitational lenses.


Is HR 5171 a Spectroscopic Binary or a Symbiotic? Is HR 5171, the largest yellow star in the galaxy a symbiotic star where one feeds off the other? Or is it a spectroscopic binary where its only possible to determine that there are two stars by spectral analysis?
Will the smaller be consumed by the larger one as was in the possible case of BP Piscium? Or is the smaller one being born?
Is the star both, one or the other or neither?
I looked at Simbad but it only calls a Double or Multiple Star System. That is too vague for me.  Calling it a Close Contact Binary is also too vague.  I know they call it an EBS.
If anyone wants an idea of the star, visit the following artist's impression.

I found arXiv:1401.2628 which says

Modeling the light curve with the NIGHTFALL program provides clear evidence that the system is a contact or possibly over-contact eclipsing binary. A total current system mass of $39^{+40}_{-22}$ solar mass and a high mass ratio $ q>10$ is inferred for the system. The low-mass companion of ${\rm HR5171 A}$ is very close to the primary star that is embedded within its dense wind. Tight constraints on the inclination and vsini of the primary are lacking, which prevents us from determining its influence precisely on the mass-loss phenomenon, but the system is probably experiencing a wind Roche-Lobe overflow.

Does the need for simulation indicate that it is a spectroscopic binary?


How would the characteristics of a habitable planet change with stars of different spectral types? The tidal forces a habitable planet experiences increase with decreasing spectral type. So a habitable planet orbiting a smaller, less massive, and cooler star would experience much stronger tidal forces than one orbiting a bigger, more massive, and hotter star, and likely be tidally locked.
But what about other factors? How do other characteristics scale up with different stars?
For example would the orbital velocity of a planet in the habitable zone of a red dwarf be smaller than the velocity of a planet in the habitable zone of a more massive and hotter star?

The following table gives the mass, radius, temperature, and luminosity of an average star of several selected spectral types:
$$\begin{array} {d|c|c|}
\text{Spectral Type} & \text{Mass} (\odot) & \text{Radius} (\odot) & \text{Temperature (K)} & \text{Luminosity} {(\odot)} \\ \hline
\text{M8V} & \text{0.082} & \text{0.111} & \text {2500} & {0.00043}\\
\hline
\text{M5V} & \text{0.16} & \text{0.199} & \text {3030} & {0.00299}\\
\hline
\text{M2V} & \text{0.44} & \text{0.434} & \text {3550} & {0.0268}\\
\hline
\text{K8V} & \text{0.59} & \text{0.587} & \text {4000} & {0.079}\\
\hline
\text{K5V} & \text{0.68} & \text{0.698} & \text {4410} & {0.165}\\
\hline
\text{K2V} & \text{0.78} & \text{0.763} & \text {5040} & {0.337}\\
\hline
\text{G8V} & \text{0.94} & \text{0.909} & \text {5490} & {0.673}\\
\hline
\text{G5V} & \text{0.98} & \text{0.982} & \text {5660} & {0.887}\\
\hline
\text{G2V} & \text{1.02} & \text{1.01} & \text {5770} & {1.014}\\
\hline
\text{F8V} & \text{1.18} & \text{1.25} & \text {6170} & {2.031}\\
\hline
\text{F5V} & \text{1.33} & \text{1.46} & \text {6510} & {3.434}\\
\hline
\text{F2V} & \text{1.44} & \text{1.61} & \text {6810} & {5.001}\\
\hline
\text{A8V} & \text{1.67} & \text{1.81} & \text {7500} & {9.3}\\
\hline
\text{A5V} & \text{1.85} & \text{1.94} & \text {8080} & {14.392}\\
\hline
\text{A2V} & \text{2.05} & \text{1.97} & \text {8840} & {21.263}\\
\hline
\end{array} $$
The following table gives the orbital distance, period, and velocity of an Earth-like planet receiving the same flux from its star as Earth does from the Sun, along with the radial velocity of the star caused by said planet and the tidal forces exerted on the planet relative to Earth:
$$\begin{array}{d|c|c|c|}
\text{Spectral Type} & \text{Orbital Distance (AU)} & \text{Orbital Period (days)} & \text{Orbital Velocity (km/s)} & \text{Radial Velocity (m/s) } & \text{Tidal Forces} {(\oplus)} \\ \hline
\text{M8V} & 0.0207 & 3.82 & 59.166 & 2.167 & 9166\\ \hline
\text{M5V} & 0.0547 & 11.68 & 50.929 & 0.956 & 977.34\\ \hline
\text{M2V} & 0.163 & 36.51 & 48.812 & 0.333 & 100.09\\ \hline
\text{K8V} & 0.281 & 70.95 & 43.134 & 0.219 & 26.5 \\ \hline
\text{K5V} & 0.406 & 114.84 & 38.518 & 0.17 & 10.11\\ \hline
\text{K2V} & 0.58 & 182.93 & 34.525 & 0.133 & 3.98\\ \hline
\text{G8V} & 0.82 & 280.06 & 31.877 & 0.101 & 1.7\\ \hline
\text{G5V} & 0.942 & 337.48 & 30.375 & 0.093 & 1.17\\ \hline
\text{G2V} & 1 & 365.56 & 29.973 & 0.088 & 1\\ \hline
\text{F8V} & 1.425 & 572.18 & 27.102 & 0.068 & 0.407\\ \hline
\text{F5V} & 1.853 & 799.11 & 25.231 & 0.057 & 0.209\\ \hline
\text{F2V} & 2.236 & 1018.01 & 23.901 & 0.049 & 0.128\\ \hline
\text {A8V} & 3.049 & 1505.21 & 22.041 & 0.039 & 0.058\\ \hline
\text{A5V} & 3.793 & 1984.29 & 20.799 & 0.033 & 0.033\\ \hline
\text{A2V} & 4.611 & 2526.01 & 19.859 & 0.029 & 0.02\\ \hline
\end{array}$$

These calculations were done with the planet being at such a distance that it recieves exactly the same flux from its star as Earth does from the Sun. In reality a planet can be significantly farther away or closer and still remain in the habitable zone.
You can see a very clear correlation: As spectral type decreases (star becomes cooler, smaller, and less massive),

Orbital distance decreases,
Orbital period decreases,
Orbital velocity increases,
Radial velocity of the star increases,
Tidal forces on the planet increase exponentially.

I'm sure there are other factors that I didn't take into account but these are the most obvious ones I could think of.

1. I got the data for the mass, radius, and temperature of each star from here. This is an average relationship and not meant to be exact.
2. The luminosity was calculated using the Stefan-Boltzmann law, assuming a perfect sphere and a blackbody radiator, which most stars approximately are.


Is the wind's intensity on Mars similar to Earth? I've read that in Mars' poles, the winds can be as fast as 400 km/h, when the poles are exposed to sunlight because the frozen $CO_2$ sublimes. I know that the Martian atmosphere is much thinner than Earth's atmosphere.

So, by knowing the wind speeds on Mars, is there any way to get an idea of its intensity, or in other words, the intensity of a wind of x speed in Mars, to which speed of wind of Earth is comparable, for them to have the same intensity?

Credit to this question for inspiration, though my calculation methods differ.
The dynamic pressure equation is $q=0.5\rho v^2$ where $q$ is the pressure, $\rho$ is the atmospheric density, and $v$ is the wind speed. If we want to know what wind speeds give us equivalent pressures on Earth and Mars, we simply generate dynamic pressure equations for each of them: $q=0.5\rho_e v_e^2$ and $q=0.5\rho_m v_m^2$, set them equal $q=0.5\rho_e v_e^2=0.5\rho_m v_m^2$, and solve for $v_e$ to get $$v_e=\sqrt{\frac{\rho_m}{\rho_e}}v_m$$ where $\rho_m=0.020 \space kg/m^3$ is the atmospheric density for Mars, $\rho_e=1.225 \space kg/m^3$ is the atmospheric density on Earth, $v_m$ is the wind speed on Mars, and $v_e$ is the equivalent wind speed on Earth.
With a velocity ratio of about 7.826 we can plug in a few values for wind speed in kilometers per hour for Mars to get:
v_mars    v_earth equivalent
   10           1.28
   50           6.39
  100          12.8
  200          25.6
  400          51.1

These could be kph, or in fact, any units of velocity. screeenshot
and here's what hat looks like in a plot:

So the 400 kph gust on Mars only has equivalent pressure of a 51 kph gust here on Earth


Are there stars in the first photo from space? Wikipedia has this photo (see below) labelled "First photo from space" which was taken in 1946 from a U.S. launched German V-2 rocket. Approximately two thirds of the image is a section of the Earth showing some clouds and their shadows. The photo is very grainy so it's hard to see much more detail than this (I can't make out any land masses as all).

In the large black section occupying the top-left corner there are some very faint dots. Are these dirt on the lens, marks on the film, JPG artifacts or are they actually stars?

I really doubt it. (Note: I'm answering this more as a photographer than an astronomer)

From this vantage point, Earth is far brighter than any stars you would see beyond it (save the Sun). When you set a camera to expose Earth, if you set it such that you can expose Earth without blowing out the details you need to set the camera accordingly. This, however, would mean that it's unlikely the camera settings would be sensitive enough to also get stars. As we can see rough details of the atmosphere in your photo (Despite the grain), it would suggest that the exposure settings would not be sensitive enough to depict stars.

For example, here is a photo of Jupiter:

There are a few things to note here:

Jupiter is entirely blown out. It's over exposed and you can't see any details of the planet.
You can see several moons. If you properly expose Jupiter, you generally won't also get the moons, because your camera settings won't be sensitive enough to pick them up. (at one time I had a more properly exposed photo that lacks the moons. Apologies, I'm not sure where it's gone off to)
You aren't seeing any stars in the background. Again, the exposure settings to pick up the moon are too sensitive to properly expose Jupiter, but not sensitive enough to pick up the stars behind it. (A brighter star, had it been in the frame would be visible in the photo, but there are certainly some stars that were in a shot this wide. Regardless, there are stars far brighter than the Galilean moons, so this isn't a particularly solid point)

There is certainly something in the photo. Judging by the amount of grain in the photo and by the fact that JPG images will always contain some artifacts it would seem more likely that what you're seeing is the affect of noise or artifacts. Dirt on the lens is unlikely - this usually doesn't show up as specks in the final photo, as this is before the light has not been distributed yet. You see dust if it's on the lens exit or between the lens and the sensor, but in that situation you're going to see it generally reducing light from the final image.

This answer may change if, by chance, the photo was taken when it was night on Earth. However, this seems unlikely for an early rocket launch and certainly for a first attempt at a photo of Earth from space.

As this is really more about analyzing a photograph than astronomy, if you want to probe further you may want to inquire in Photography.SE.


Spectroscopy question about a Paschen series absorption I had this question on a test and got it wrong but I can't understand the explanation.

I don't get why we receive a Paschen series absorption. Could anyone explain?

The Paschen series absorption requires a population of H atoms in the $n=3$ level and a radiation field containing infrared photons $\leq 1875$ nm, that can boost them in to the $=4,5,6...$ levels.

If hydrogen gas is at 25,000K, you can use the Saha equation to work out the fractional ionisation. This depends on both the temperature and density of the gas. For the temperature you have been given, the ratio of ionised to netutral hydrogen is $1.7\times 10^{25}/n_e$, where $n_e$ is the electron number density in m$^{-3}$,

The typical particle number densities of an interstellar gas cloud are much, much less (many orders of magnitude) than $10^{25}$ m$^{-3}$, so the gas is totally ionised.

If that is the case then there can be no Paschen absorption and no Balmer absorption due to the cloud since there are no atoms in the $n=2,3$ states. There may be small amounts of Paschen and Balmer emission from the cloud via recombination.

Thus what you basically see is the spectrum of the star and since you aren't told what kind of star it is, then it's impossible to rule out the presence of emission lines or absorption lines of any kind. It is thus a dreadful and unanswerable question.


How does dark energy remain constant? According to cosmological constant, dark energy should have a constant density but since the universe is expanding so is the volume of the universe and so should the mass of dark energy.

But how does dark energy gain mass or energy?

Not all models of dark energy assume it constant.

However, when this is assumed true, dark energy is usually compared to the energy density of the vacuum. In other words, while the Universe expands, the dark energy increases proportionally, so its ratio with the Universe volume (the density) stays constant.

Or, better, we generally believe those models where the dark energy is a constant because they fit better our observations of the Universe, in particular the Big Bang model, which describes how our Universe evolves.

If you like some math, you can have it here.


Why does the sun appear 13x bigger through diffraction in Voyager 1 image from 1990, but not from Earth? The sun in the Voyager 1 photo here from 1990 appears 13x bigger than its diameter, diffraction of course. On Earth, photographs of the sun do not make it appear 13x bigger. Why?

A description of the photo from jpl.nasa.gov,

  "The wide-angle was taken with the camera's darkest filter (a methane
  absorption band), and the shortest possible exposure (5 thousandths of
  a second) to avoid saturating the camera's vidicon tube with scattered
  sunlight. "

Don't base the size of the Sun on what you see in that image because the Sun is over-exposed.  The big spot in the center of the diffraction spikes are saturated pixels because they all appear to have the same value (the spot looks uniform) so you don't know what their real values are; you would need to take a shorter-exposure image (or put filters in front of the lens) if you wanted to take a proper picture of the Sun.  Since Venus is a lot dimmer than the Sun, they set the exposure long enough to see it and accepted the fact that the Sun would be over-exposed.  You see the same thing in some of the Hubble images where to bring out the background stars, the brighter ones are over-exposed.

Another way of looking at it, if you see an image where you can very clearly see the diffraction spikes, that part of the image is over-exposed.


Calculate time when star is above altitude 30° To find the best observation time for an object, I'd like to calculate the time when it is 30° or more above the horizon. Local Sideral Time would be sufficient.

To include that in my program, I need the formula.

Example:
On June 4th, Jupiter has the coordinates
RA= 9h 19m 28.0s
Dekl= 16° 32' 0"

It rises at 10:32 and sets at 00:05.

After rise, when is it at altitude 30°, and after transit, when is it at altitude 30° again ?

I found this formula at http://www.stjarnhimlen.se/comp/riset.html. Although it's for the sun, it seems to be what I'm looking for.

$$\cos (\text{LHA}) = \frac{\sin (\text{h}) - \sin (\text{lat}) \times \sin (\text{Decl})}{\cos (\text{lat}) \times \cos (\text{Decl})}$$

Applied to the sample assuming a latitude of 45° I get.

Is this the correct approach?

Yes, this is the correct approach.  The $h$ in the equation is the altitude above the horizon of the object at which you consider it to be rising or setting.  This is typically non-zero, because of atmospheric refraction, and, in the case of the Sun or the Moon, because of their finite diameters.  In your case, the object 'rises' when it climbes above $h=30^\circ$ and 'sets' when it drops below that altitude.

If $\left|\cos(LHA)\right|>1$, there is no solution, because your object never crosses the $h=30^\circ$ line.  $LHA$ is the local hour angle, and you can find the local sidereal time $\theta$ using

$LHA = \theta - \alpha,$

where $\alpha$ is the right ascension of your object.


Is the shape of observable universe and shape of space same? I have read that Euclidean space is considered as flat space and Minkowski space is flat space-time. So when we say our observable universe is flat are we saying that the space is flat or space-time is flat? I know the Planck data has proved the space to be flat but is it only space or space-time? Since in general relativity matter curves space-time and not just space.

When talking about curvature in (General) Relativity one usually refers to the metric interval and how this depends on the space-time coordinates.

For example, the metric interval in Minkowski space can be written
$$ ds^2 = c^2\ dt^2 - dr^2$$

This is an example of both flat space and flat space-time. It is flat space because if I want to find the distance between two events happening at the same time $t$, it is simply the integral of $dr$.

Similarly I can ask what the time interval is between two events and this is just given by the integral of $dt$.

A counterexample is seen when using the Schwarzschild metric, which is appropriate in the space outside a spherically symmetric mass. Considering only the radial and time components, the metric is
$$ds^2 = c^2 (1 - r_s/r)\ dt^2 - (1- r_s/r)^{-1}\ dr^2,$$
where $r_s = 2GM/c^2$ is the Schwarzschild radius associated with the mass $M$.

Now there is no simple relationship between $dr$ and the spatial separation between two events that occur at the same $t$, we must do an integral. Similarly, to work out the time interval between two events at the same $r$ we must do a more complicated integral. Both space and space-time are curved.

The evolution of the universe as a whole is governed by the Robertson-Walker metric.
$$ds^2 = c^2\ dt^2 - a(t)^2\frac{dr^2}{1 - kr^2},$$
where $k$ is a curvature parameter that may be -1, +1 or zero.

At the moment, the best measurements are consistent with a zero curvature - that the universe is spatially flat. That does not mean it is flat on much large scales than we can observe - that is still possible. A curved universe that was exponentially "inflated" shortly after the big bang may well mimic a flat universe even though the curvature is finite.

Note though that because the universe is inherently "lumpy", this spatial flatness is only true on scales much larger than planets, stars, galaxies and even clusters of galaxies.

Whether you could call space-time flat in a spatially flat Robertson-Walker universe I doubt. The reason for this is the time-dependent scale factor $a(t)$, which ensures there is a coupling between the $r$ and $t$ coordinates. It is this coupling that is responsible, for example, for the cosmological redshift of light.

A technical answer would need to look at the Einstein Tensor which characterises the curvature of space-time caused by mass and energy. In the Robertson-Walker metric, this tensor definitely has non-zero components, so you would have to say that even if space is flat, space-time is curved.


Collapse of a star below the mass of Chandrashekhar Limit How is Chandrashekhar Limit calculated? What happens to stars below the mass of the Chandrashekhar limit after they collapse?

The Chandrasekhar mass is a theoretical maximum stellar mass that can be supported by ideal electron degeneracy pressure.

It was originally calculated by Chandrasekhar using polytropic equations of state for white dwarf stars of various compositions.

It is found that if the equation of state is $P \propto \rho^{4/3}$, which is appropriate for ultra relativistic electrons, then equilibrium is only obtained for $M_{CH} = 5.8 \mu_e^{2} M_{\odot}$, where the density becomes infinite and $\mu_e$ is the number of atomic mass units per electron in the gas.

Typical white dwarfs are carbon and oxygen with $\mu_e= 2$, and $M_{CH}= 1.45M_{\odot}$, but in the core of a supernova progenitor made of iron-56 it would be lower - around $M_{CH}=1.25M_{\odot}$.

In more recent years, the Chandrasekhar limit has evolved to colloquially mean the maximum possible mass for a white dwarf. The main corrections to the ideal case considered by Chandrasekhar are Coulomb corrections, the possible onset of inverse beta decay (electron capture) and the instability at a finite density predicted by using GR rather than Newtonian gravity. The latter probably sets the "Chandrasekhar mass" for carbon white dwarfs to be $1.38M_{\odot}$.

A carbon white dwarf that had a mass less than this could be supported by electron degeneracy pressure and would not collapse.

If your question is asking for a derivation of the Chandrasekhar mass, then I suggest you look at the stability analyses that are presented in a standard text on the subject like "Black holes, white dwarfs and neutron stars" by Shapiro and Teukolsky. An approximate result can be obtained using the virial theorem assuming a uniform density star.


Appearance (frequency) ranking of stars? Is there some ranking / statistics that shows how common given type of celestial objects is? A table that would allow the answers to questions like: What type of star is most common per volume of our galaxy? Are white dwarfs more common than red giants? What percentage in the number of celestial objects located are the black holes? Pulsars? How many red giants have been catalogued?

In general smaller stars are more common than larger, more massive stars.
This is typically described by the Initial Mass Function which describes the number of stars of a given mass which form initially. Moreover, larger, more massive stars burn through hydrogen quickly and live shorter lives, making them even less common as time passes. Thus, to determine the distribution of objects of various types of equal masses, you need to start with an initial mass function, and then use a stellar evolution model to evolve the system to the time of interest.

More generally, a Universal Mass Function, is an attempt to answer the question you pose in terms of mass. That is, are more massive or less massive objects more common in the universe?

The Universal Mass Function provides a description the density of occurrence of objects of various masses per cubic parsec, shown in Figure 3 of Kroupa et al. 2011. As you can see, the more massive something is, the less common its occurrence. Thus, black holes which form from massive stars will be far less common that small low mass stars which become white dwarfs. There are still uncertainties the exact distribution, shown at least partially by the shaded portions of the figure, and it will vary from one region of space to the next.

For more info check out Kroupa et al. and Binggeli and Hascher. 2007. “Is There a Universal Mass Function?”.

Kroupa, Pavel, Carsten Weidner, Jan Pflamm-Altenburg, Ingo Thies, Joerg Dabringhausen, Michael Marks, and Thomas Maschberger. 2011. “The Stellar and Sub-Stellar IMF of Simple and Composite Populations”. ArXiv e-print 1112.3340.


What is the present day radius of the observable universe? The radius of the observable universe is generally taken to be ~45 billion light  years. However, we see distant galaxies as they were many millions of years in the past, so there are two ways I can interpret this radius value:

1) Light that is reaching Earth for the first time today was emitted by ancient galaxies that were initially a lot closer, but are now estimated to be 45 billion light years away from us.

2) Light that is reaching Earth for the first time today was emitted by ancient galaxies that were initially 45 billion light years away, and are now a lot further.

Which interpretation, if they're not both wrong, is correct?

The answer is basically #1. The observable universe contains objects whose light has taken up to the age of the universe to travel to us.

Thanks to the expansion of the universe, the things that emitted that light are about 46 billion light years away now.
The exact figure depends on the details of the cosmological parameters and also whether you take the first time that the light can have started out as being the "epoch of recombination" (when the cosmic microwave background was emitted and the universe became transparent to radiation) or the big-bang itself (which would add about 0.5 billion light years to the figure).

Light emitted just after the epoch of recombination has a redshift of about 1100. That means the scale factor has increased by a similar amount. Thus something that is now 46 billion light years away was about 42 million light years away when the light was emitted (of course neither the Earth or our Galaxy existed then).

The answer cannot be #2 since light from something 45 billion light years away cannot have reached us in the age of the universe.


Problem with Dobsonian reflector telescope I'm not sure whether I should ask this question here but I'll give it a shot. I have a XT8 Dobsonian Reflector telescope and it works fine. But recently when I use the focuser to focus the image when I change eyepieces it focuses fine but it makes a scratching noise that is really loud and it doesn't sound right. Now I don't know what this screw that I am about to explain is neccesarly for because I am no where near a telescope expert. Next to the eyepiece there is a screw that you can turn loose and if you make it as tight as possible the focus knob won't focus the image but when it is a little loose it starts to focus it, but it keeps making a scratching sound and a high pitched rubbing sound that doesn't sound right.  I know I explained this pretty vaguly but any help would be helpful! So far I could not find anything on the web for this weird problem.

UPDATE:

Next to the eyepiece there is a screw that you can turn loose and if
  you make it as tight as possible the focus knob won't focus the image
  but when it is a little loose it starts to focus it

Sounds like the lock screw. If that's what it is (your description is a bit vague), it's designed to lock the focuser in place, like when you're doing prime focus photography with a heavy camera, and you don't want the focuser to move around and lose focus.

If it's the lock screw, don't mess with it. Don't tighten it, you don't need that for regular observations. Keep it comfortably loose. If it's a bit tight, it could be rubbing against the drawtube (or whatever moving part it's designed to stop), and then it would make that noise - and you don't want that.

I have a very heavy 30mm 82deg Explore Scientific eyepiece, and even with that thing (bigger than a soda can, and weighing about 1 kilogram) I don't need the lock screw - and this is on a regular mass produced two-speed focuser, not some fancy precision hand-crafted MoonLite. Visual observations are all about moving the focuser around, hunting for the best focus, and changing eyepieces for the best view. Let the focuser move easily.

Unless it's moving too easily, but then you have a problem with the focuser.


Configuration required to tidally heat Earth-like rogue planet Could it be possible for Earth-like rogue planet to maintain large bodies of liquid water on the surface without a sun by relying mostly on tidal forces of a moon (or multiple moons).
How big/far/fast would the moon have to be?

What about a binary planet instead of planet and a moon?

Do you need tidal heat?  What about core heat?

Ice is a fair insulator.  Put enough ice on top and the heat flux from the core can keep the bottom of the oceans liquid, no tides required.

If you need the water on the surface, I don't think it's possible. Even a huge amount of interior heat (whether from tides, radiation, or primordial) wouldn't be sufficient to hold the surface liquid.  The Earth possibly had periods where there was no surface water (snowball earth), and that was with the advantages of an atmosphere and solar input.

The lack of a sun is going to make having an Earth-like atmosphere difficult.  In an interstellar environment, any thin atmosphere will freeze out.  Without an atmospheric blanket, the surface temperature plummets.

So I think you could have an ocean, but not with surface water.

One more thing.  Assuming the current heat production of the earth, and its size, I calculate that something around 5.6km of ice is sufficient insulation.  Water below that depth with average heat flow would be liquid (ignoring pressure effects and only assuming temperature and heat flow issues).  That's a lot of ice, but not implausible.


How did New Horizons take such well-lit pictures of Pluto? The photos of Pluto from New Horizons are truly beautiful.

But considering that Pluto is so far away from its nearest start - our Sun - how is it so well lit up?

Did the New Horizons have a massive flash mounted on its camera to illuminate the planet or is there just a lot of ambient light in space?

I'm really curious.

It's brighter on Pluto than you think.

NASA developed a tool called Pluto time, which tells you when at your place the ambient light conditions are similar to the ones on Pluto.  This occurs when the Sun is only 2° below the horizon!  That's quite shortly after sunset, and considerably before the end of civil twilight, which is when it's 6° below.

All of these photos were taken at local "Pluto time":

Pluto time, according to NASA.  Source: NASA

To answer your question: all it takes is a slightly longer exposure time / larger aperture / higher gain (what photographers may call ISO) than taking photos closer to the Sun.  It's easily bright enough for outdoor activities (except perhaps on a cloudy evening and/or in a dense forest, but neither should be common beyond the orbit of Neptune), so have fun glacier hiking on Pluto!


How can ionized emission line flux decrease as a function of increasing metallicity or abundance? The chemical evolution of galaxies is an important way to learn about their formation and stellar/gaseous constituents. Many galaxies show narrow emission lines at optical wavelengths (3500-9000 Angstroms) from ionized elements (e.g., [O II] at 3727 Angstroms) and recombination Balmer lines (e.g., H$\alpha$ at 6563 Angstroms). The origin of the ionized emission line flux can depend on the amount of the element present in the galaxy, and also on the nature of the ionizing source (i.e., the oft-mentioned "shape of the ionizing spectrum").

I have a few questions that I cannot seem to find clear answers to in textbooks, publications and review papers. My main reference for the following facts/questions is Kewley & Dopita 2002, ApJS, 142.

What is the difference between metallicity and abundance, and what impact would these two parameters have on the ionized emission line flux from a galaxy (assuming that the ionizing spectrum is kept constant)?
There are claims that some ionized lines (e.g., the auroral line [O III] at 4363 Angstroms) would become much weaker as the metallicity increases. Why? I would think that higher metallicity means more Oxygen and, assuming we keep the ionization spectrum fixed, more likely for more [O III] to exist. Should this not increase the [O III] 4363AA flux?
Similarly, some excitation line ratios (e.g., ([O II] 3727 + [O III] 5007)/H$\beta$, commonly known as $R_{23}$) are known to decrease with increasing abundance (measured via log(O/H)+12). Again, why? I would think if the metallicity or oxygen abundance is higher, then there is an increased chance of getting more [O II] and [O III] emission. Also, assuming that metallicity is higher, the amount of hydrogen should decrease, and this would decrease the $H\beta$ flux in the $R_{23}$ ratio above, thus increasing the ratio at high abundance.

Metallicity and abundance

Metallicity

Without specifying a given metal, the term "metallicity" — abbreviated $Z$ — usually refers to the total metallicity of all elements, i.e. the mass fraction of all metals to the total mass of some ensemble of elements, e.g. a star, a cloud of gas, a galaxy, etc. (as usual, the term "metal" refers to all elements that are not hydrogen or helium). For instance, the mass of all metals in the Sun, divided by the Sun's mass, is 0.02:
$$
Z_\odot \equiv \frac{M_\mathrm{C} + M_\mathrm{N} + M_\mathrm{O} + \ldots}{M_\odot} = 0.02.
$$

Sometimes we talks about the metallicity of a given element, e.g. oxygen. The mass fraction of oxygen in the Sun is 0.005 (i.e. oxygen comprises 1/4 of all metals by mass), so we say $Z_\mathrm{O} = 0.005$.

Unfortunately it is not uncommon to implicitly talk about the metallicity of an object, divided by Solar metallicity, such that a galaxy which has one-tenth of Sun's metallicity is said to have $Z=0.1$, rather than $Z=0.002$.

Abundance

The term "abundance" is only used for a single element. It basically expresses the same thing as metallicity, and is often used interchangeably, but is expressed in terms of the number $N$ of element nuclei, and as the ratio not to all nuclei but to hydrogen nuclei. For wacky historical reasons, we also take the logarithm and add a factor of 12. Taking again oxygen as an example, the mass fraction of 0.005 corresponds to a nuclei fraction of roughly $5\times10^{-4}$, so we say that the abundance of oxygen is (e.g. Grevesse (2009))
$$
A(\mathrm{O}) \equiv \log \left( \frac{N_\mathrm{O}}{N_\mathrm{H}} \right) + 12 = 8.7.
$$

Metallicity of a given species vs. total metallicity

In general, the ratio of a given element to all metals is roughly constant. That is, various elements are produced by stars approximately by the same amount. But various processes may cause elements to exist in various forms. For instance, metals deplete to dust, but some elements tend not to form dust, e.g. Zn. For this reason, Zn is a better proxy of the total metallicity than, e.g. Mg, since half of the Mg may be locked up in dust.

Metals increases cooling

Elements also appear in various excitation states, which depend on various processes. The lines you mention, [O II] and [O III], arise from collisionally ionized oxygen, which subsequently recombines (in my first answer I wrote, wrongly, that it was excited), and thus depend on the temperature of the gas. As the metallicity of the gas in a galaxy increases, the ratio of the intensity of these lines to that of hydrogen lines (e.g. H$\beta$) first increases, as expected. However, the increased metallicity also allows the gas to cool more efficiently. The reason is that metals have many levels through which electron can
"cascade" down.
If the electron recombines to the level where it was before, a photon of the same energy will be emitted, which itself may radiatively ionize another atom. But the many levels in metals makes de-excitation to intermediate level more probable, such that the electron cascades down, emitting several low-energy (infrared) photons, which are incapable of ionizing atoms and thus escape. The result is that energy leaves the system, i.e. the system is cooled.

This in turn means that, above a certain metallicity threshold — which is specific to a given species — the abundance of the collisionally excited lines begin to decrease. The following figure is taken from Stasińska (2002), and shows the turnover for the two oxygen lines:

This means that measuring the metallicity of a single species in general gives two solutions for the total metallicity. Luckily, as the turnover is different for different elements, measuring the metallicity for several species can constrain the total metallicity.


How would seasons at the equator be if the Earth were tilted 90 degrees? If Earth's axis of rotation were tilted 90 degrees, what type of seasons would be observed at the equator.

My prediction is that there would be no seasons, as the orientation of the equator would be the same as if the Earth

I assume you mean "what if the Earth's equator was perpendicular to it's orbit?" in which case the seasons would be greatly exaggerated at the poles and day and night would only really mean something for about half the year if you're nearer the equator. Temperature differential would be much more significant than it is now.

Unless you mean that the axis of Earth's rotation is exactly perpendicular with respect to it's orbit, in which case there would be no seasonal variation whatsoever. (Not including Malkovitch cycles and solar cycles)


What's the object in the sky? Days before, I saw this pic. I don't know what it is.

Edit: This pic was taken by a Taoist at an ancient ruins in Mongolia on 2012-07-15 04:12 UTC+8 (day break). He said it looked strange.

Edit: I installed Stellarium, changed the time to 2012-07-15 04:12 UTC+8, and faced east. It seems to be the Moon.

The photo shows a conjunction of the Moon and Jupiter.
A few hours later, observers in Europe
(APOD,
virtualtelescope.eu)
saw the Moon pass directly in front of Jupiter, an event called an occultation.

Stellarium shows Venus and Aldebaran below them, but in the photo they are probably behind a cloud.
Capella is visible at far left, slightly higher than the Moon.


How do I get flux density when I have mJy/beam , beam size and pixel size in an astronomical image I have a fits file of the galaxy m87 produced by SCUBA 2 and I am wondering how I obtain the fluxes at 450 and 850 microns. the output image gives pixel magnitudes in mJy/beam where I want to find eh flux for a certain part of the galaxy. I am as far as summit the values of all pixels within an aperture but I am not sure where to go from there to obtain the flux as there are more than one pixel within the FWHM. say I have a an aperture of radius 1.5 pixels that gives a sum of 6985.04885203 mJy/beam with a pixel size of 3.99999999 arc seconds and a beam size of 13 arc seconds how do I obtain the flux for that aperture at that wavelength?

In essence read of the peak flux value and if a point source this will be the same as your integrated flux.

to elaborate on this:

In this question your source appears to be comparable to your beam. so I would treat as a point source. - note you might also want to consider that many people apply a "match filter" to their data if dealing with faint point source emission - this smooths the map - increasing the effective beam size within the map and also there may be some flux loss (see for example: http://adsabs.harvard.edu/cgi-bin/bib_query?arXiv:1707.00990)

You will see this (mJy/beam to mJy) is also discussed here: http://www.eaobservatory.org/jcmt/help/

For a real point source, a peak brightness value reported in units of mJy is the same as a peak brightness value reported in mJy/beam.

But what happens if we have a map in mJy/beam and we want to obtain an integrated intensity value, a total flux value? We first sum up a number of pixels and now we want to get our units correct from mJy/beam to mJy…

Total Flux = flux summed over a number of pixels/(number of pixels in a beam)

Then your units are:

[mJy*pixels/beam] / [pixels/beam] = [mJy].

Now we know that For a Gaussian:

Beam Area =  2 × π × σ² [arcsec]

where the σ of the Gaussian beam can be calculated from the JCMT FWHM values at 850 and 450 microns (reminder the beam components are provided in Dempsey’s 2013 paper).

FWHM = 2 σ √(2 ln 2)   [arcsec]

So we can use the FWHM to obtain σ to calculate the Beam Area and report the beam area in terms of pixels:

number of pixels in a beam = Beam Area [arcsec] / (pixel length)²

Finally I should add you will also want to do an aperture correction - see Dempsey et al 2013's calibration paper if you are using an aperture of a different size than 60".


How does the Sun orbit the Sun-Jupiter Barycenter? I got the Orbital Elements of Jupiter around the Sun, which describe how it orbits relative to the "fix" Sun.

Jupiter
Semi Major Axis (AU): 5.20336301

Eccentricity: 0.04839266

Inclination to Ecliptic Plane (deg): 1.30530

Longitude of the Ascending Node (deg): 100.55615

Longitude of Perihelion (deg): 14.75385

Mean Anomaly at J2000.0 (deg): 19.65053

I want to convert this One-Body Problem back into the Two-Body Problem to find out where the Sun is gonna be at a given time (i couldnt find the Orbital Elemens of Sun around Sun-Jupiter Barycenter).

So the Semi-Major Axis depends on the Mass Ratio. The Eccentricity and Inclination should be equal for both, just like Longitude of the Ascending Node. The Longitude of Perihelion should be the one from Jupiter+180°, doesnt it?

The main Problem is the Mean Anomaly at J2000.0. The two Ellipses are equal, besides of the Size. Thus, the True Anomaly at a given time should be the same for Each Coordinate System. Then the eccentric Anomaly will be the same too. Does that mean the Mean Anomaly for J2000.0 will be also the same for both?

Thanks for reading and sunny greetings!!

The barycenter of two point masses M1 and m2 -- or their 'center of gravity' (which to avoid any doubt is not in itself a center of gravitational attraction) -- lies on the straight line joining them, and divides the distance between them in the proportion m2:M1 -- in classical/Newtonian terms. That is, the barycenter is separated from center M1 by the fraction m2/(M1+m2) of the total distance M1 to m2, and by the complementary fraction M1/(M1+m2) from the center of m2.

In a two-body system, the positions and motions of each body relative to their barycenter are similar in the way that two triangles can be similar: the corresponding angles for each are the same, the distances are in a constant proportion. (See also https://en.wikipedia.org/wiki/Barycenter and https://en.wikipedia.org/wiki/Two-body_problem .)

So at least part of the answer to the question about the Sun and Jupiter -- imagined as a 2-body system -- is that the angles including the mean anomaly are indeed the same for each.

However, this only applies approximately to the figures quoted in the question. These are not 'one-body' or 'two-body' orbit figures, they are heliocentric approximate mean elements for Jupiter in the n-body solar system as at J2000 (i.e. 2000 Jan 1.5), from a derivation applied to data over the interval 1800-2050. The figures appear in a GSFC/NASA Jupiter fact-sheet, and also on p.316 of 'Explanatory Supplement to the Astronomical Almanac' (1992, ed. K P Seidelmann et al.), where the error in Jupiter's position computed from the elements is stated as up to 300" in right ascension.

Mean elements such as those quoted in the question can be derived from the time-varying osculating elements obtained over a long period. Such elements may be obtained for the chosen period in different ways, for example

(a) by deriving them from results of a long-term combined numerical integration of the orbits of a set of solar-system bodies -- usually involving all of the major planets along with either some of the minor planets as further individual bodies and/or an approximate allowance made for the effects of those not included individually;  or

(b) by deriving them from an analytical theory (a method used for example to obtain the results in J L Simon et al., 'Precession formulae and mean elements for the Moon and the planets', Astronomy & Astrophysics 282 (1994) 663-683).

In either of these cases, subtraction of the periodic terms leaves the mean elements as non-periodic polynomials as a function of time-interval from a chosen epoch associated with the data.

A third and approximate approach, which seems to have been the one adopted for the figures quoted in the question, involves fitting the osculating elements to a chosen polynomial function over a chosen interval. The website http://www.met.rdg.ac.uk/~ross/Astronomy/Planets.html suggests that the figures here resulted from a 250-year least squares fit of data from the JPL DE200 numerically integrated planetary ephemeris to a Keplerian orbit where each element was allowed to vary linearly with time.  The mean elements for a date such as J2000 would then represent the value of the linear function at the chosen date.

If it is wished to take the approximate mean elements for Jupiter at J2000 to represent 1-body elements on that date, then that is perhaps about as close an approximation of the kind as can be made: and the angles will then be similar as already described, and the barycentric distances will be obtained by the fractional split already mentioned.

A more precise way of tracking the relation between the Sun and the solar-system barycenter is outlined in a 1983 paper from JPL on solar-system numerical integration by XX Newhall et al.). This takes into account all of the planets and their masses, and the result is a complex path that has perhaps never been represented by any trigonometric series, but only numerically and diagrammatically -- e.g. at https://en.wikipedia.org/wiki/File:Solar_system_barycenter.svg, thus:


Which solar system objects have the least stable solar orbits? My research is not giving me anything definitive. Are the Apollo asteroids orbits less stable than the Centaurs or any Trojans, or is another class of objects in less stable solar orbits? On what timescale are those orbits unstable?

The least stable orbits are likely the temporarily captured orbiters of Earth and other planets. These are bodies that have been captured by the Earth/Moon gravity well and move from solar orbits to terrestrial orbits. They normally have complex orbits that take them well beyond the orbit of the Moon and most don't last long.
Most such objects are small, typically less than 1 m in diameter. There are probably about 1000 such bodies with sizes of 10cm and up orbiting the Earth. Normally, they are undetected but occasionally a larger object will be discovered.  Most recently, 2020 CD3 was captured in the winter of 2015-16 and escaped Earth's gravity in May 2020.
The orbits of such temporary orbiters are not stable, they normally escape the Earth/Moon gravitational well after a few years, and return to heliocentric orbits.


Earth orbit around the Sun and its position in the Galaxy Given the orbit of the Earth around the Sun and the position of the Sun in the Galaxy, in which season are we (the Earth) closer to the galactic center?

The radio source Sgr A* is at RA 17h46m Dec -29°, or about 6° south of the ecliptic at longitude 267°.
The Sun appears most nearly opposite that position, ecliptic longitude 87°, about three days before the June solstice.
At that time of year, the brightest part of the Milky Way is highest in the sky around local solar midnight.

However, the distance from here to Sgr A* is about 7.9 kpc or 1.6 billion au.
The Earth is only infinitesimally closer to the galactic center in June than in December.


Multiple apparent lunar eclipses last night I saw something very strange last night that I'm hoping someone can help me explain.
I was sitting on my roof with a night-cap at about 3am last night. The moon, about the last quarter, was low in the sky in the east. Something truly strange happened. The moon started to disappear. There were clouds in the sky, but it wasn't this, I'm quite sure. The moon was being eclipsed. The dark part (at the top of the moon, as the moon was lying on its back - a sailor moon? That's what it looked like, like you could sit in it.) grew, the crescent closed, until it was just a spot of cream colored light and then winked out entirely.

I was a bit amazed - I'd stumbled onto my roof in the dead of the night and by complete chance witnessed a total lunar eclipse? I sat there in awe, staring. That's when the moon started to reappear, in the reverse procedure as its disappearance. Becoming a gleaming point of light, then a thin crescent, then the quarter moon which I suppose is what it is "supposed to look like."

AND THEN IT HAPPENED AGAIN. The moon was eclipsed and revealed again.

Both times the whole process took about 6-10 minutes. I'm not sure if this cycling continued, it was late and I had to sleep. I checked the NASA site for eclipses for my time and location, but no.

Has anyone ever seen anything like this, or have any plausible explanation as to what is going on here?

This was clouds.

Eclipses occur at full moon, not last quarter.
Eclipses take several hours as the moon traverses the Earth's shadow, not a few minutes.
Eclipses don't look like the moon becoming a gleaming point of light. In a lunar eclipse the moon becomes a copper colour, as it is illuminated by red light that has passed round the Earth, through our atmosphere.

On the other hand, clouds can pass in front of the moon at any time, and when low in the sky would take a few minutes to pass in front of the moon (depending on wind), and could entirely cover the moon, giving a the impression of an eclipse, if you've not seen one before. But lunar eclipses look like this:

Photo Credit: Jimmy Westlake space.com


Are the stars in all galaxies orbitting around a middle? In our Milkyway all the stars are orbitting around the black hole in the center. But what is the cause it this rotation and are there galaxies where the stars aren't orbitting a center?

Stars aren't really orbiting the center of a galaxy, as much as they're orbiting the gravitational potential of the galaxy. The galaxy doesn't need to have a black hole in the center for the stars to orbit, and even if it does (which in fact it does most of the time), the black hole does not dominate the gravitational potential, except in the most central part (see sphere of influence of a black hole).

All matter in the galaxy contributes to the gravitational potential, but since dark matter comprises 5/6 of the mass, this is the most important contributor. Only in the hypothetical case of a perfectly symmetrical distribution of masses, would the black hole and the center of rotation coincide. And since the paths of the stars are ellipses and not perfect circles, the center of the potential will not lie in the center of their orbits, but in one of the focal points. Furthermore, galaxies are dynamical and the potential continually changes, in turn changing the exact orbit of a star.

Galaxies form due to the collapse of a dark matter halo, and the gas that follows  along. Such a halo will in general have a non-zero angular momentum, which is partially conserved during the collapse (they do lose a significant amount, for instance through minor and major merging; see e.g. D'Onghia et al. 2006).

As in the case of a forming star, a forming planet, an ice skater, or you on an office chair, reducing the radius means increasing orbital speed. This is the origin of the rotation.


What was the "brilliant new star in Aquila" on June 8, 1918, just after the solar eclipse? This great answer about the US Naval Observatory's $3,500 expedition to Baker City Oregon to observe the June 8, 1918 total solar eclipse links to the January 1919 Popular Astronomy article about the expedition. The article notes on page 6 the following coincidental observation:

  An interesting occurrence was the appearance of the brilliant new star in Aquila. This was seen by Mr. Peters and Mr. Conrad on the night of June 8 and the discovery was communicated by telegram to various observatories in the country. It had however been discovered earlier on the same night by other astronomers in the east.

What is the story behind this bright new star in Aquila? What was it? Who discovered it first?

It was Nova Aquilae 1918.  Wikipedia has some detail. https://en.wikipedia.org/wiki/V603_Aquilae
Apparently it was discovered by Zygmunt Laskowski, an amateur astronomer.


Plot of best available resolution vs wavelength - radio through gamma rays? What I'm looking for is a graphic that shows in a general way the best available telescope resolution vs wavelength throughout the entire wavelength spectrum. So for example, there might be two very high resolution peaks aroud

millimeter wavelengths (ALMA)
visible wavelengths (HST and many ground telescopes with Adaptive Optics)

The beautiful image below from this great answer got me thinking. I have reposted it from there.

  Image courtesy of Wikipedia user Hunster under the Creative Commons Attribution-Share Alike 3.0 Unported license.

  The infrared, ultraviolet, and x-ray images come from the Spitzer Space Telescope, the SWIFT observatory, and the Chandra observatory, respectively.

After a bit of searching, I found this blog page, which has several charts about various observatories, including this one:

Image courtesy of Olaf Frohn under the Creative Commons Attribution-Share Alike 4.0 License.

The majority are space-based, although the radio telescopes are largely land-based. They cover existing and future telescopes, at energies from the gamma-ray spectrum to radio waves. You are correct, too, in assuming that adaptive optics can cause dramatic increases in angular resolution; CHARA and the European Extremely Large Telescope both use adaptive optics, and actually can have better angular resolutions than some space-based telescopes.

I annotated the graph to cover in green the smallest angular resolution at various wavelengths:

Notice that most of the lines in the radio, microwave, and infrared part of the spectrum are diagonal, with roughly the same slope. This is because they are limited by diffraction. In the case of radio waves, this is because the atmosphere has little impact. In the case of infrared- and visible- wavelength telescopes in space - and in space-based telescopes in general, the main thing that stops them is the diffraction limit.

The diffraction limit is
$$d=\frac{\lambda}{2n\sin\theta}$$
where $\lambda$ is wavelength and $n\sin\theta$ is the numerical aperture. On a log-log plot, such as the one above, we have
$$\log d=\log\lambda-\log(2n\sin\theta)$$
and
$$\frac{\mathrm{d}\log d}{\mathrm{d}\log\lambda}=1$$
for all telescopes limited by the equation. Thus, telescopes restricted by this limit should be described by a diagonal line with a slope of 1 (-1 on this graph).


What is the difference between aurorae and electroglow? One of the discoveries of Voyager 2 at Uranus was a phenomenon called "electroglow", which as I understand it is related to charged particles interacting with the atmosphere that cause the emission of ultraviolet radiation. What I am not clear on is how this process is different to the production of aurorae. Is there some clear difference between the two processes that allows them to be distinguished?

Aurorae are caused by charged particles (usually electrons) being drawn into the polar atmosphere by the planet's magnetic field.
Airglow is caused by just photons interacting with the atmosphere.
Electroglow requires both electrons and photons.
Source: The Free Library  (article is under express copyright, see the paragraph that starts "Voyager 2")


Greenhouse effect on planets Venus has a greenhouse affect because its atmosphere is 96 percent carbon dioxide. However, although mercury is closer to the sun it does not have the greenhouse effect because it has no atmosphere and only an exosphere. I was wondering why doesn't mars have a greenhouse effect like Venus? They both have around 95 percent carbon dioxide in their atmosphere and consist of an atmosphere. In addition, I was also wondering about what are the requirements to have a greenhouse effect on a planet?

For there to be a significant greenhouse effect there has to be a lot of greenhouse gas in the atmosphere. Not just a large percentage, but a large amount of gas

There is a greenhouse effect on Mars, however, as its atmosphere is very thin. The surface pressure on Mars is only 0.6% of the pressure on Earth. As such there very little of anything to trap heat. Even though a large percentage of the atmosphere is CO2, the greenhouse effect only raises temperatures by about 5 Kelvin. http://nova.stanford.edu/projects/mod-x/id-green.html

It wasn't always this way. During a dust storm, more heat can be trapped, and temporarily raise temperatures. Early in Mars's past the atmosphere was thicker at temperatures were higher. It is even possible that at one point, a runaway greenhouse effect heated Mars, until the atmosphere thinned and the planet cooled http://www.astrobio.net/mars/did-ancient-mars-have-a-runaway-greenhouse/

The only requirement for a planet to be warmed by the greenhouse effect is for its atmosphere to contain greenhouse gases. Mars does have a greenhouse effect, as other bodies with atmospheres: Earth, Titan and Venus. https://www.nasa.gov/home/hqnews/1991/91-143.txt Other rocky bodies don't have a thick enough atmosphere for a greenhouse effect to be measured.


how does Kepler stop exposure after a 5 minuate or 30 minute cadence time? There are two kinds of cadence time for Kepler, 5 minutes and 30 minutes.
I just know there is no shutter and the CCDs of Kepler can be locked after exposure. handbook

But how to realize that? Remember we have to not affect the photometry.

The other question is so far how many unique sources has Kepler observed? 160,000?

They don't stop. The Kepler CCDs read out the signals collected during the  accumulated time of 6.02 seconds. The fixed read out time is 0.52 seconds. So each CCD gets one frame every (6.02 + 0.52) = 6.54 seconds. Then Kepler sums up every 9 frames (short cadence) and 270 frames (long cadence). The time between two short cadences is (6.54 x 9) = 58.9 seconds, and (6.54 x 270) = 1766 seconds between two long cadences, but the exposure time is (6.02 x 9) = 54.2 seconds for a short cadence, and (6.02 x 270) = 1625 seconds for a long cadence.

There are totally 95 million pixels available on Kepler CCDs, but the spacecraft can store only 5.4 million. The maximum number of targets is about 170,000. see http://archive.stsci.edu/mast_faq.php?mission=KEPLER


How did the solar system get its angular momentum? Why did Earth not fall into the Sun at the start? I know that Earth is always falling towards the Sun, but due to its sideways momentum it just misses always.  But how did it all start?  To go into detail, when the gas cloud with zero angular momentum collided, our solar system was made with planets and sun having their own angular momentum which makes them rotate.  But how did the revolution start in the beginning?  Were  they placed in such a position and distance that they started attracting each other but how ? .They were having only rotation .  But a revolution started which forced them to follow a elleptical path  so how this happened in the beginning?  Did some external force act on them and revolution started?  How was  this revolution started?

i have assumed dust cloud was having zero angular momentum but it can have any value that does not affect the question i think ?

A particle's angular momentum around a point is the particle's mass times its distance from the point times its velocity around the point.  So a small velocity at a large distance can still mean a lot of angular momentum.

The bit of a Giant Molecular Cloud which collapses to form one solar system is typically on the order of a light-year across or somewhat less. It normally possesses some angular momentum -- just from random motions if nothing else.

When it collapses to form a solar system, its size drops from maybe a light-year down to maybe 10-100 AUs. A light year is about 60,000 AUs (astronomical units), so it is collapsing to about 1/1000 the size. To conserve angular momentum, the circular velocity of the matter must increase by a factor of a thousand.  And the fraction of the matter which will form the star collapses by another factor of a thousand or so.

Not only does this give you plenty of speed, but for the star to form it actually needs to shed angular momentum.

See the Wikipedia article on angular momentum for a lot of good details.


Are there other pictures of Titan surface from Huygens? I can't find any other picture of the Titan surface from the Huygens probe than this one

Are there any? Are they not disclosed on purpose?

If this is the only one taken, why?

Titan surface http://i.picresize.com/images/2016/11/17/uXaDQ.jpg

One second of googling reveals the whole archive:

http://esamultimedia.esa.int/docs/titanraw/index.htm
(note you can click onto the strips to inspect them!)
The archive depicts the whole decent of the Huygens lander onto Titan, with the first picture being on the ground, and some of them seem to be taken from higher up altitudes, pages 17-27 you can see the flow deltas)
The three pictures per strip show the views of the three cameras into different directions.

However the location on the planet doesn't change, just the viewing angle around the spacecraft upon descent. This is because Huygens was a lander without any additional propulsion, as it was already considered to be a massive feat if the lander would make it to the surface at all.
Also a propulsion system would have been more expensive, and more or less useless, as the expected lifetime of the lander in Titan's atmosphere was a few hours, which happened to be true.

So summa summarum, from the final position in which the lander was fixed, yes this is the only picture of Titan's surface.


Is it possible that Mercury was originally the moon of Venus after a giant impact? Mercury looks like the Moon, and so it makes me think about a question: is it possible that Venus and Mercury were the a same planet originally, and a giant impact with that planet made it split into Mercury and Venus (like with the Moon and Earth)?

This was originally going to be a comment, but it ran too long, so I'm making it an answer.

Some models argue that the scenario of a satellite of Venus escaping like this is unlikely. Alemi & Stevenson (2006) have explored the possibility of a prior Venusian moon, starting from the assumption that Venus would not have been able to avoid a giant impact. Here's their sequence of events:

A large body collides with Venus in a similar manner to the proposed Earth-Theia collision.
Debris from the impact moves outwards into a disk surrounding Venus,
A moon coalesces from the disk, and begins to slowly recede because of tidal acceleration.
Another large body hits Venus. It reduces Venus's angular momentum, reversing its rotation.
The moon spirals into Venus as it undergoes tidal deceleration, finally colliding with it again.

One of the tricky things about testing this model is that the authors say that there would not necessarily have been drastic composition changes, meaning that it would be hard to analyze the planet's surface and see if there is evidence supporting the double impact hypothesis. So far, there have not been tests.

It is certainly true that Venus could have suffered other impacts - the model does not preclude that. There are a couple problems with Mercury arising from such a collision:

Other impacts could have ended up with the same result as the original moon.
The chances of many more impacts aren't too high.
Solar tides would likely have destabilized the orbit of any moon larger than a few kilometers in diameter (see Sheppard & Trujillo (2009)).
MESSENGER determined that Mercury has a high potassium/thorium ratio on its surface, which would seem to disprove any events involving extremely high temperatures, including any giant impact variant.

Of course, if we accept that Venus could have captured a moon, only the third objection remains - still a strong point against the survival of a satellite, even by itself.


What is this constellation? Every night I see this constellation from my window but I can't identify it. What is this constellation called? Also, it would be helpful if you could clarify on the other stars as well.
Actual photo:

Pattern:

You can see the constellation Orion on this photo:

(Screenshot from Stellarium)
Aligning the original photo with this screenshot, I found the highlighted stars to be the following:


What projection would I use to map the northern night sky onto a geodesic dome? I'm trying to create a geodesic dome with the stars of the northern night sky inscribed on the outside. The highest point of the dome would be the celestial north pole. How do I create a map that projects onto the "net" of the geodesic dome, such that when I assemble the dome, the stars are in the correct positions? I have an equirectangular map that I have rotated 90 degrees north so it is centered on the pole.

My existing map is a high-resolution jpeg file in an equirectangular projection. I have some experience creating and manipulating panoramic images, but much less experience converting between map projections! I would like to use a computer program to convert this projection into a printable template if possible.

The mathematically ideal scheme would be a separate gnomonic projection
(see Wikipedia or
IMO)
for each tile of the dome.
Many planetarium programs implement this projection as an option.
Fortunately, each tile boundary is a straight line on both adjoining maps.
Unfortunately, continuity across tile boundaries requires some care in choosing each tile's scale and center of projection.


Why can there be several seconds of lag between a binary neutron star merger and the emission of gamma rays from the same area? Answers to: Gravitational waves and gamma ray burst: how were the error bars determined for this speed of gravity calculation? Was $H_0$ used? go into substantial detail explaining this determination. These have been helpful, which encouraged me to try to read the linked paper again. However, based on comments and the paper itself, I see that assigning limits to what sounds to me to be an emission lag of several seconds between the end of the merger and a subsequent gamma ray emission from the same general object/area had to be at least considered and included in some way.
Here I'd like to ask only about this lag. While I'd also like to know what produces the time-lagged gamma ray emission I'm not sure a priori if that is a different question and/or if it's been answered already.
Question: Why can there be several seconds of lag between a binary neutron star merger and the emission of gamma rays from the same region?

The paper (section 5.1) discusses three possibilities in the context of a relativistic fireball model, where some of the kinetic energy in relativistic jets of material emerging from the explosion is converted into gamma rays. The gravitational wave emission is always "prompt" since any surrounding material is transparent to gravitational waves. In contrast, the gamma ray emission may be delayed because:

The gamma rays are produced by shocks. There is a delay in the jets reaching the radius where the shocks form, either generated internally in the fireball by components with different speeds or just through deceleration in the interstellar medium.

The fireball is initially opaque to gamma rays and it is not until it has expanded sufficiently that it becomes transparent, allowing the gamma rays to escape.

It may take some time for the jet to punch out of the dense inner regions of the explosion.


Where can I get J2000 epoch initial mean longitudes for all planets and celestial bodies? I'm following this paper to calculate geocentric coordinates. The authors have provided the initial mean longitude values for the planets (provided in attached table) for  01.01.2000 @ 00.00 UT but I'd like Uranus, Neptune, Pluto and some asteroids also. Is there anywhere this information is freely available or did they calculate it? Many thanks.

Take a look at this link. It included Program Codes Listing also (Appendix C). But if you only want the initial mean longitude and the angular veclocity... the appendix C.1 does have them listed at the beginning... in it, the values are L0 and n for each planet... (The n in the listing is in arcsecond, you have to divide it bye 3600 to get your desired degree value.)


What is the speed of time? From the theory of relativity, if any object is moving with the speed of light or more than the speed of light, then the time stops or seemed to be stopped for that object. That means, that the speed of time for an object moving with the speed of light is zero.

So what will be the speed of time for an object in the space, travelling at a negligible speed(i.e.,zero) or approximately stationary.

The expression "speed of time" doesn't make much sense, since speed is the rate of change of position with time, and "time" doesn't have a position. Hence it doesn't have a speed.

If you want to know the amount of time dilation at a given speed, it is answered by special relativity. $t' = t\gamma $ where $t$ and $t'$ are the amount of time that passes in the two frames, and
$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}} $$

At low speeds, $\gamma$ is approximately 1, and at high speeds it approaches $\infty$.

To fit with your notion of the speed of time as being the relative rate of passing of time in the two frames, an expression for the speed of time would be $t/t' = 1/\gamma$, which is close to zero at high speeds, and close to 1 for low speeds.

So the speed of time at low speed is equal to 1.


Neutrinos passing through black holes What would happen to a neutrino that, traveling parallel to the black hole, crossed it's event horizon?  Would it come through the other side unaffected? Or, A neutrino and a black hole just happen to be traveling in the same direction but the neutrino get close enough to cross the event horizon.  Is it now trapped inside the event horizon even though technically it only interacted with the black hole on a gravitational level?

In general, don't think of black holes as material objects like stars or planets. We can ascribe properties like mass, charge and angular momentum to them, but if you're near a black hole and reach out to touch its event horizon, you won't feel anything (assuming you're still alive). Your hand will simply pass through empty space, and then into the black hole. Think of black holes instead as regions of spacetime, with the curious property that if you enter one, you can never exit.

I'm guessing your question is motivated by the fact that neutrinos interact extremely weakly with normal matter. They have very little mass and no electric charge, and mainly influence other particles through the weak nuclear force - which is, of course, weak, although not as weak as gravity. The problem with taking this assumption further and concluding that a neutrino might pass through a black hole unharmed is that black holes aren't "normal matter". They're sections of spacetime, and while they may contain matter, that matter is not at all important when interacting with other particles. The only thing here that is important is that they are sections of space which have only one-way openings. Once inside, nothing can leave.


How do you estimate the error on the height/width of a Gaussian? I'm trying to fit Gaussians to several lines in a spectrum that I have. Some of them overlap with one another, causing the fitting program that I'm using to not be able to give reasonable estimates for the errors on the measurements. For example, sometimes it will give the width of the spectral line to be 0.5 GHz but give the uncertainty at 1,000 GHz.

To fix this, I've been doing manual estimates on the ones that it can't get. I read that you can estimate the error on the peak of the Gaussian with:

Peak Error = (1/2)*(Width / Signal To Noise Ratio)

but I can't find anything for the height or width. Are there similar ways to estimate the uncertainty on these measurements?

Thank you.

If you know the number of overlapping Gauss functions, you may write them as $\sum_{i=1}^n a_ie^{-(\frac{x-\mu_i}{\sigma_i})^2}$, then subtract your time series, with more than $3i$ items, to reduce the risk of an ill-posed problem.
Then apply a root mean square (RMS) minimization / least squares method to infer the $3i$ parameters.
If your FWHM, or $\sigma$ is known to be the same for all peaks, your number of parameters reduces to $2i+1$.

See also Rietveld refinement as a related approach.


When are gravitational waves emitted during a merger of black holes? When black holes are merging, and emitting gravitational waves, are we seeing those waves start when the event horizons are merging, or when the singularities are merging?

The gravitational waves are actually emitted as long as you have a quadrupole moment with a second time derivative that is non-zero. The quadrupole is defined as:
$$ Q_{ij}^{tt}(x) =\int\rho \left(x^i x^j - \frac{1}{3}\delta_{ij}r^2\right)\mathrm{d}^3x.$$
To simplify this, here is a rule of thumb: if you have a system that is not spherically symmetric and that is changing with time (like two black holes), you can expect gravitational waves to be emitted. For example, the system made of the Earth and the Sun is emitting ridiculously small gravitational waves (I think it is emitting a few Joules per year), but it is.

Because of how hard the waves are to measure, you can only expect to see them on Earth for black holes in their last stage, just before they merge, when the emission of gravitational waves is at its peak.

For the LIGO detection, the beginning of the signal as observed by our detectors was in the inspiral phase. This is the phase when the black holes are getting closer one to each other and losing gravitational energy (through the emission of gravitational waves), before their horizons touch. The end of the phase is actually when the black holes get within their so-called ISCO (Innermost Stable Circular Orbit), which is the smallest radius a massless particle could orbit in a stable fashion. Anything inside this orbit will slowly spiral inward and fall into the black hole.