How to locate nebulas visible through AstroMaster 114 telescope without device assistance? I will be going to the desert this month and it's my understanding that the night sky will be very dark there. I've got some practice locating the planets but I'd really like to get a look at a nebula or galaxy. I understand from Is the Celestron astromaster 130 EQ Model able to see nebulas and galaxies? that nebulas and galaxies are visible with modest magnification. I'd really appreciate some tips on locating stuff that's cool to look at other than planets and the moon. I've got an AstroMaster 114 (114mm, f/8.77) with a couple of eyepieces and a Barlow lens.

I've searched a bit and these targets seem like a good short list.

Albireo
Orion Nebula, M42
Andromeda Galaxy, M31
Hercules Cluster, M13

I understand that my telescope has an equatorial mount and I do take care to orient it northwards and adjust it to match my latitude, but I've never truly bothered to center it on the north star. I find locating the north star a bit daunting. All the instructions I've encountered say to locate the big dipper (aka the plow) and follow the line of two if its stars. Finding constellations is not my strong suit -- especially with the haze and obstructions here in Los Angeles. I can usually find Orion pretty well, though.

I have an app on my iphone called Planets which I find quite helpful for locating the planets, but I'm not sure if it will operate properly out there where there's no telephone service. It also doesn't provide any tools to identify or locate deep space objects. It's just for planets.

My telescope does have some "setting circles" (little dials demarcated in degrees or hours, see photo) but my understanding of ascension and declination is sketchy at best and I also understand that these are not especially accurate. I'm wondering if they might have some use to get me in the ball park of the objects I'm seeking -- maybe help me at least find the constellation from which I can locate the deep space objects of interest.

So to summarize, I'm hoping for tips on locating the north star and establishing a proper baseline position for my telescope and I'm also hoping for good, basic instructions for locating some interesting, non-planetary objects to view in my scope.

Those targets are good choices for beginners at the right times of year.
However, in June M42 is a daytime object, and M31 rises in the wee hours, leaving only M13 and Albireo in the evening.
Other good summer targets include:

M8, the Lagoon Nebula
M11, the Wild Duck Cluster
M57, the Ring Nebula

Precise polar alignment is only required if you plan to take long exposure photographs.
Setting circles help when you can't see much of the sky at once, e.g. under an observatory dome or partial cloud cover.
In order to be useful, the right ascension circle must be reset each observing session, or more frequently if your mount isn't motorized.

The Skymaps.com Evening Sky Map can help you find your way around the night sky.
It shows the whole sky as it appears 2-3 hours after sunset, including several deep-sky objects visible with a 100mm or larger telescope.
A multi-page atlas such as the S&T Pocket Sky Atlas provides enough detail to actually get those objects in the telescope's field of view.
A little red light for the maps minimizes the impact on your night vision.

With practice, you can learn to locate any target of interest by triangulation, first among the bright stars, then among the faint stars in between.
For example, M13 is about 60% of the way from Arcturus to Vega and 30% of the way from η Her to ζ Her.
A detailed atlas also shows two 7th magnitude stars each ¼ degree away from M13 in an obtuse triangle.
If the object does not at first appear in the telescope field, carefully sweep the surrounding area.
Some mounts provide fine motion controls to help with this.


What's still needed before we can observe orbits of exomoons thereby weighing exoplanets? Comments below this answer to How do we weigh a planet? point out that we currently cannot (or at least have not) detect moons around exoplanets, much less measure the sizes and periods of their orbits around their planets.
I commented there that:

...detecting exoplanets wasn't possible, nor did direct imaging of planetary systems, nor did making parallax measurements from the Kuiper belt, nor was the detection of gravitational waves...

my point being that impossibly challenging observations keep getting done.
So I'd like to ask what the most likely extension of observational Astronomy would be that would allow for the measurement of the period of exomoons thereby weighing exoplanets.
Is it simply a pair of space telescopes in a halo orbit with laser interferometry between them and plenty of time? Could it be done from Earth? After all, exoplanetary systems are already imaged from Earth.

My opinion (and I think there has to be a large element of that) is that the presence, and ultimately identification of the orbital period, of exomoons is going to come from very precise transit photometry.
If a sufficiently large moon orbits a planet, then this will leave its signature in the transit light curve. A "Fourier-type analysis" of the light curve might then reveal a periodic nature to the light curve shape during transit that could be attributed to the orbit of the moon. The requirements here would be a sufficiently large moon that its relative position alters the transit shape. The exoplanet also needs to be in a short period orbit so that you can observe lots of transits.
A related (and better) technique, that in practice would be used simultaneously, would be to look for the signature of a moon using changes in the duration and timing of transits that recur in a periodic way. The exoplanet-exomoon barycentre is what follows a Keplerian orbit; but the area-weighted "centre of opacity" of the system will, in general, not coincide with this barycentre because mass is proportional to $r^3$, while obscuring area is proportional to $r^2$. This will lead to a wobbling of both the time of  transit
and the transit duration that may again yield a periodic signal which could be identified with the exomoon period. These methods are discussed extensively by Kipping 2009, who point out that the transit timing and transit duration signatures have different dependencies on the exomoon mass and separation from the exoplanet and thus a careful measurement of both could yield the exomoon mass. i.e. Each method in its own has a degeneracy which means the exomoon mass/period could not be determined, but measured together, the degeneracy is broken.
There is a trade-off here. A wide exoplanet-exomoon separation will give bigger signatures, but of course by Kepler's third law, the orbital period will be longer and so you would likely need a longer dataset of transits to identify any periodicity and pin down the amplitude of these signatures.
At this stage, I don't think there is any likelihood of hearing about an exomoon period/mass in the next few years. Exomoon detection by these methods is though a possibility and may already have occurred (see Teachey & Kipping 2018) . Perhaps the PLATO mission, which will produce better light curves than TESS or Kepler, and have long datasets will stand a better chance (and indeed exomoon detection is one of the mission goals - see Rauer et al. 2014).


Is the delay to correlate the signal from one dish to another on a VLA radio telescope longer for red shifted objects? When observing 2 objects side by side close to the horizon with a very large array, one receding fast & red shifted, is the delay to correlate the signal from the closest dishes to the further ones longer for the red shifted object?

...is the delay to correlate the signal from the closest dishes to the further ones longer for the red shifted object?

Good question!

The short answer is no. The path difference between the closest and farthest dish in the array comes from local geometry, a big triangle in the air right above the VLA dishes. Here the speed of light is essentially constant (except for local temperature and humidity gradients in the air above the array, or maybe wind sheer) and so wouldn't vary between the to signals very much for reasons related to the sources' Doppler shifts.

The interference path length difference is divided by the local speed of light (in the air) to get the time difference needed for best correlation of those two dishes, and it doesn't make any difference what the source of the signal is or the speed that it was moving.

Doppler shifted photons don't look or act any different than non-shifted photons, there's no way to tell the difference in an experiment, unless you have some information about the original frequency before the shift.

The only teeny tiny difference might be if the air has a different index of refraction at the shifted frequency than the unshifted frequency. That's certainly taken into account during normal analysis, and I don't think it rises to the level and intent of your question.


Are there any planned future observations of TRAPPIST-1 and its planets? The recent discovery of new planets around the star TRAPPIST-1 happened by observations from 7 ground-based telescopes, as well as the Spitzer Space Telescope. The team of astronomers responsible has noted that there are many relevant parameters which still have large uncertainties, including orbital information and planetary mass. There's also some excitement about possible atmospheric composition.

Are there any definite planned future observations of the system? I assume that at some point it will be targeted by the James Webb Space Telescope, but that won't launch for another year and a half, and it has plenty of other targets.

I'm not looking for detailed information - I know that cannot be easily gathered - but if there are any general mentions, like this one about the Subaru Telescope and Planet Nine, that would be fantastic.

A website trappist.one has been created to detail information about this discovery. In the "Future" section, they specifically state:

In the short term, photometric follow-up using the repurposed Kepler satellite (named K2) along with with newer observations using Spitzer ought to reveal the period of planet 1h. We will also search for additional planets. Theses lightcurves, combined with ground-based measurements will increase the number of transit timing measurements for each of the planets. This will give us more accurate masses and orbital eccentricities. This will confirm whether the planets are mostly rocky or whether they contain a certain amount of volatiles, like water.
In the medium term, we can expect the first attempts at detecting the atmospheres of the TRAPPIST-1 planets, using Hubble, followed by deeper investigations thanks to the James Webb. The James Webb could in principle measure the temperature of the planets, and detect the chemical composition of their atmospheres. It will do so by collecting dozens of eclipses of the TRAPPIST-1 planets. The advantage of having seven planets in one system is that we will be able to compare them to one another.

It seems they're currently using Kepler and Spitzer and plan to use Hubble in the near future as well as JWST after it launches.


If the Earth is rotating counter clockwise, how does the sun rise from east? I was very confused when I watched this video from 2:18 till 2:45. And it seems very clear that if the Earth is really rotating from west to east, shouldn't be the Sun rising from west and setting in east? We all know certainly that the sun rises from east, but how come Earth is rotating in the way as shown in the linked video?

PS: I somehow feel very stupid for asking this question...

Try this non-Gedanken-but-real experiment:

Hold an apple in your hand so the Sun (or a light) shines on it. Nerds may use a proper globe.
Along the apple equator draw arrows to the right with an E for East: E→E→E→E, .... Nerds be sure to use a non-permanent marker.
Spin it counterclockwise (when looked at from above, i.e. the North pole). You're doing it right when all the Es move in the direction of the arrows. West to East.
The Sun rises on all points on the apple that go from dark to bright; identify that line. It should be half a circle.
Imagine a near indestructible humanoid killer machine from the future standing on that line (called terminator, the line, that is). What direction does the machine look when it glances at the Sun the moment he (the Sun, that is) rises?


Looking for methods to preserve night vision Does anyone have a method to preserve night vision so that a car or other passing light source does not ruin it?
I have a dark location to observe the sky; however, it is near a rarely used dirt road.  I can hear a car coming in time to protect my night vision, but have been unable to find a method so that its light does not ruin my light vision.   I have tried putting a 5-gallon metal bucket over my head, covering my eyes with a towel, etc.   My night vision still gets ruined.
I want to be able to totally block all light from my eyes until the car/light-source passes.

Your problems with shielding your eyes from light probably come because it is easy to shield them from direct light from the car lights, but hard to shield them from reflected car light coming from every angle.

I suggest that you experiment with closeing your eyelids while shining lights at your face.

I just tried shutting my eyes while looking at a light about a foot and a half from my face.  Everything looked a very dark grey  or black color, and it didn't look any brighter when I looked at the light bulb.
I often close my eyes when a passenger in a car and facing toward bright sunlight.  The sunlight shines red through my eyelids and I tend to see geometric patterns - when the car passes into shade from trees the color changes to blue.  So the geometric patterns sometimes flash between a red and a blue background rapidly.
But looking at the light bulb right now I don't see any light shining through my eyelids, because sunlight is many times as intense as the light bulb.
So possibly you  might want to experiment with staring at lights of varying intensity with you eyes closed to see how much light, if any, penetrates your eyelids. And yu can try calculating how bright a car's lightswill be at your distance from the road, and comparing it to light sources which doo or do not penetrate your eyelids.
Maybe simply closing your eyes is enough to prevent loosing your night vision.

You could wear a helmet with a hinged opaque face plate that you can slam down when you hear a car coming.  You could have opague cloth handing down several feet from the bottom of the helmet to prevent light from getting in through the bottom, or maybe tape the bottom of the helmet to your clothing.

If you own the place you observe from, you could built a tiny windowless shed or shack there and make certain it is pitch black inside with the door closed, and run into it and close the door when you hear a car coming.

You could set up a tent made of totally opague fabric or plastic sheets, and when you hear a car coming jump into it and knock down the tent pole to collapse it on you.  When the car passes, struggle to get out of the tent, perhaps by finding the tentpole and putting back up.

I hope you might find some of my suggestions useful.


Why are stars still distant from one another? Why are stars so far from each other? Shouldn't gravity pull them closer over time? And if the effects of gravity are negligible is there an explanation why stars have to be so distant from one another?
The closest star is Alpha Centauri (I think) and it is 4.4 light years away.

The initial star formation regions were regions that have a high enough mass density to form a star. The density of the early universe was not constant at different locations. Some regions had high enough density to form a star, and some didn't.
When a star forms it draws in matter from a large distance away. This forms an accretion disk and leaves a temporary relatively empty space around the star at a large distance away. There is no way a star can form in this low mass region. Once you get further away from this region the matter density may return to a level where a star can form. But this region is far from a star.
Yes, gravity from a particular star is pulling at other stars but there are many stars pulling on the particular star of interest. So the net force on the star may be very small.


When will the number of stars be a maximum? There are very roughly a "mole" of stars in the universe. Wikipedia quotes an estimate of $3 \times 10^{23}$ though the number is associated with some debate and uncertainly.

I'd like to know if there are estimates of when the number of stars in the universe will maximize. Is it expected to increase asymptotically to some maximum, or will it peak and then decrease.

I suppose this could depend on what the definition of "star" is taken to be, if brown or black dwarf objects are counted or not. I don't want to pre-specify, it's more likely that a good, well-informed answer will include this information.

TL; DR Somewhere between now and a few hundred billion years time. (For a co-moving volume) Now read on.

If stellar remnants are included, then the answer is very far in the future indeed, if and when the constituents of baryons begin to decay. So let's assume  that "stars" means those things that are undergoing nuclear fusion reactions to power their luminosity. Let's further assume that the stellar mass function, ($N(m)$ is the number of stars per unit mass) we see in the neighbourhood of the Sun is representative of populations in all galaxies at all times (difficult to make a start, without assuming this).

The number of stars that have been born is equal to the sum over time (the integral) and over mass of $N(m)$ multiplied by the rate at which mass is turned into stars in a comoving volume of the universe $\Phi(t)$.

We then need to subtract a sum over time and mass of the rate of stellar death in the same comoving volume. The rate of stellar death is the rate of stellar birth at a time $t-\tau(m)$, where $\tau(m)$ is the mass-dependent stellar lifetime. We ignore mass transfer in binary systems and assume that multiples can be treated as independent stellar components.

Thus the number of stars at time $t$ is approximately
$$N_*(t) = \int_0^{t} \int_m N(m) \Phi(t') - N(m)\Phi(t'-\tau(m))\ dm\ dt'\ .$$
To find where this is a maximum, we differentiate with respect to time and then equate to zero. i.e. We look for the time when the stellar birth and death rates are the same.

I was going to (and possibly still will) attempt some sort of analytic approximation, but Madau & Dickinson (2014) have done it better and taken into account the metallicity dependence of stellar lifetimes and the chemical evolution of galaxies. The star formation rate peaked about 10 billion years ago, is more than an order of magnitude lower now and is exponentially decreasing with a time constant of 3.9 billion years.

The integrated stellar mass is shown in their Fig 11 (shown below). It is still increasing today, but at a very low rate and has not passed through a maximum. The reason for this is that most stars have masses of 0.2-0.3 solar masses and lifetimes much longer than the age of the universe. Even if these stars are added at a very slow rate, their death rate is zero at present.

If star formation did continue at a low-level then the number of stars would only begin to significantly diminish once the stars near the peak of the stellar mass function, that were born at the earliest times, start to die. The lifetime of a 0.25 solar mass star is around a trillion years (Laughlin et al. 1997).

On the other hand if star formation ceased now then the number of stars would immediately begin to diminish.

Perhaps we could argue that the current exponential decline will continue and the peak
will come in another few billion years when stars of 0.8-0.9 solar masses begin dying off. However, that is futurology given that we have no first principles theory that explains the time-dependence of star formation, so I believe the best answer that can be given is somewhere between now and a few hundred billion years time.

Note that this answer assumes a co-moving volume. If the question asked is phrased in terms of the observable universe then because the number of stars has nearly reached a plateau, then the answer becomes close to whatever age the volume of the observable universe is maximised. I say "close to" because you have to factor in that the observable universe includes stars in distance slices at all cosmic epochs. I am unwilling to undertake this horrendous calculation, but note that the current concordance cosmological model has our observable universe slowly increasing from around a radius of 45 billion light years now, to about 60 billion light years in the far future Davis & Lineweaver 2005, and this may compensate for a slow decline in the number of stars in a co-moving volume.


Can we see the Big Bang happen if we look far enough? The observable universe is constantly expanding as more light from the Big Bang reaches us. This light has been travelling for billions of years, so we are looking at the universe as it was a few billion years ago. This being said, would it be possible to look at the Big Bang in action by looking deep into the universe?

No. The furthest we can see is the cosmic microwave background radiation (CMB). Early on (after the big bang), matter was fully ionised and the electrons frequently interacted with the photons. That has two consequences. First, the radition was that of a blackbody at the same temperature as the matter. Second, the universe was opaque, i.e. photons couldn't travel very far.
Due to the explosive expansion of the universe, the temperature decreased all the time until eventually atoms formed. This is called the epoch of re-combination, though the "re" makes little sense. At that point the universe became suddenly transparent and we can see most radiation emitted ever since. In particular the radiation field from the epoch of recombination has been redshifted and appears to us as the CMB.

edit
To answer the speculation in the comments what if the universe was not opaque. Well, all we would be able to so is some (nearly) isotropic radiation field emitted very early on (and redshifted by the cosmic expansion since the big bang). The CMB is exactly that, but emitted at redshift $z\sim$1000 (the epoch of re-combination) rather than larger $z$.


How would Jupiter's brightness relative to our sun seem to a remote observer (observing from a remote star) Given that Jupiter is 5 AU from our sun and a remote observer viewing our solar system from some other part of our galaxy looks at it. We assume that Jupiter's radius is $11 \times 6700$ km.

What would be the relative brightness (luminosity?) of Jupiter relative to our sun's?

The absolute visual magnitude of the Sun is about 4.8. (This is the Sun's brightness viewed from 10 parsecs).

The brightest visual magnitude for Jupiter is -2.7, when it is about 4 au from the Earth. Using the usual formula we can therefore calculate the absolute magnitude for Jupiter as 25.9. (Note that this is way brighter than the faintest objects seen for instance with the Hubble Space Telescope; but as Jupiter would be separated from the Sun by only 0.5 arcseconds, a telescope with even better angular resolution than the HST would be needed to pick it out of the glare from the Sun).

Thus to first order Jupiter is 21 visual magnitudes fainter than the Sun (a factor of 250 million) . The situation is slightly worse than this because as seen from outside the solar system, Jupiter could not be fully illuminated, because it must have an angular separation from the Sun to be visible at all. A half-illuminated Jupiter would be at least a magnitude fainter (actually a factor $\pi$ for a Lambertian reflector).

But all is not lost. If we we to switch to the infrared, the situation is far more favourable. The approximate ratio of Jupiter/Sun brightness improves from about $1.4 \times10^{-9}$ in the visible to about $2.8\times10^{-8}$ at a wavelength of 10 microns (Traub & Oppenheimer (2010). But this is still a 19 magnitude difference.

The technology to achieve these sorts of contrasts is not quite there yet, even if we were viewing from nearby stars.  An example of what is currently possible can be seen with the discovery of a few Jupiter mass planet around the star GJ 504 by Kuzuhara et al. (2013). The star-planet contrast was about 15-16 magnitudes at near-infrared wavelengths of 1-4 microns and with a star-planet separation of 2.5 arcseconds (equivalent to Jupiter as seen from only 2 parcsecs).


Why doesn't this paradox disprove (some) multiverse quantum gravitational theories? As I understand it, one theory of the multiverse is that there are an infinite number of universes separated by small distances in other (than our observable 3/4) dimensions and that gravity is weak because it permeates these other dimensions.

But if there are an infinite number of these universes then surely the gravitational force we would feel would be infinite?

The answer to your question for brane cosmology is simply that gravity follows the inverse-square law. We don't feel the gravitational pull of, say, Mars because it's so far away. Any other universes are so unimaginably (all right, that's hyperbole) far away in comparison that any gravitational effects by the matter in them would be very, very, very, tiny.

A similar argument might be to say that an infinite series always goes to infinity because you keep adding on terms. But this is often not the case, as with the series
$$\sum_{n=0}^{\infty}2^{-n}=2$$
It is infinite, but it converges.


Which class of objects in Solar System carries more mass? Could you help me sorting this list in terms of total mass carried by these classes of objects?

The star (Sun) - it's very big but only one.
The planets (incl. dwarf planets). Certainly beaten by the Sun, but...
the asteroids, meteorites, comets, all that rubble. They aren't heavy individually but there's a lot of them.
the interplanetary medium. It's very low density, but it overshadows the rest by far in terms of volume occupied.

I'm not sure where to attach the moons - some of them are larger than some dwarf planets, some are definitely more asteroid-sized. I guess - whoever decides to answer this, is free to pick where they'd like to include the moons (or maybe to split them between the two classes, say, attaching the ones with hydrostatic equilibrium to dwarf planets, and bundling the rest with the rubble)?

The sun by far occupies the largest part of the solar system's mass. The mass of the sun happens to be approximately $(1.98855±0.00025)×10^{30}$ kg. Which is immensely huge. By comparison the inter-planetary medium though may occupy a large volume is nowhere dense enough to even compare to the mass of the sun which is approximately 99% of the solar system's mass.

The mass of all planets combined by comparison happens to be approximately $2.6634×10^{27}$kg.
As far as interplanetary matter is concerned, its density is to low and its volume is pretty hard to define so mass remains a highly variable quantity.

Moons and asteroids account only for a fraction of the mass of the solar system though they happen to be present in large numbers.


60 second exposure photo with phone at eyepiece with an app produced giant ball of light, what next? Last night I tried my first long exposure photo of M44 (the beehive cluster). While my exposure time was 60 seconds, my photos came out with a giant ball of light in the center of the photo.

Could this be due to my set up, as I only was using a phone app and putting my phone up to the eyepiece for one minute?

It's hard to tell, however the very likely solution of your problem lies in a combination of two or three effects:

The image is not focused. Did you focus on bright stars or the moon and not change the phone afterwards? This a common problem and (I personally) find it hard with a phone to get a sharp image through a telescope. On the telescop I have available, I need to hold my phone about 1cm away from the eye piece to get a sharp image. The focus WILL need to be different than for you looking through the telescope with your naked eye. An un-focused image will be just that: bright area where like arrives and black outside, showing vignetting.
the image is motion-blurred.
you really took too long an exposure with too high an ISO rating. Reduce either and see whether you get better results. At highest ISO rating my DSLR camera yields bright white images after a comparable time (I didn't try my phone but suspect similar sensitivity behaviour). How long exactly depends on how dark your local sky is.

From all these points the most important is: make sure you have your camera-telescope-combination focused. Test that on bright objects, thus a sharp moon or smallest-possible dot for a bright(!) star imaged. You can do the focusing in video mode and highest ISO setting. When that is ensured, then only start to image interesting objects.


Can a bullet be fired on the moon and sent it into orbit? I know you need oxygen for the ignition, but presumably, if the bullet is impervious (or water-tight) and if there is a little air enclosed in the bullet case, wouldn't that be sufficient to fire it ?

And, giving the speed, would that be enough/too much to send it into orbit around the moon ? Would you need to fire it at a certain angle ?(which will presumably make it an elliptic orbit)

Almost. (EDIT: actually yes, see below)

The speed of an object in orbit depends on the radius of the orbit, and the mass of the body being orbited.

The lowest orbit possible is where the bullet is just grazing the surface, so r = the radius of the Moon. M is the mass of the Moon, and G is the gravitational constant. Let's do the math:

wolfram alpha calculation

The result is approx 1.7 km/s.

A .50 BMG Saboted Light Armor Penetrator (SLAP) reaches 1.2 km/s muzzle velocity. So even a very powerful round such as this one moves a bit too slowly.

It is conceivable that a tricked-out .50 round, with an extra load of powder, and possibly shot from a specially prepared rifle, could reach the 1.7 km/s speed needed; that would be a 40% increase in velocity over the SLAP round.

EDIT: According to this article saboted .224 caliber bullets loaded into .300 RUM (.300 Remington Ultra Magnum) cases can reach a speed of 1.7 km/s (the average they measured over the whole test was a little less than that, but the highest values exceeded 1.7).

In that case, if fired at the horizon, it would do a circular orbit, would go round the Moon and hit the shooter in the back 1 hour and 47 minutes later, provided it doesn't hit a mountain or something en route, and provided the orbit is along the equator.

If you're not shooting along the equator, it will still do an orbit, but its motion relative to the surface of the Moon will be a weird spiral and it will not return to the starting point (the orbit itself will still be a closed loop, a circle, but the Moon spins underneath it, so seen from the Moon it appears like a spiral).

The circular orbit (standing on a hill, shooting at the horizon) requires the smallest muzzle speed to complete the loop. Any other orbit would require a greater muzzle velocity from the rifle. The more elongated the orbit, the higher the velocity.

Plus, if you shoot higher than the horizon, the bullet would hit the ground before it reaches you. You have to stand on an elevated point (a hill), and make sure the rifle does not point at the ground either forward or backward. This is because the return trajectory of the bullet after completing one loop is aligned with the rifle (the rifle's barrel is tangent to the trajectory), and if the butt of your rifle points at the ground, on the return trajectory the bullet would have to come out of the ground, which is impossible.

A laser mounted on the rifle must not point at the ground no matter whether the laser is pointing forward or backward, otherwise the orbit gets clipped by the ground.

TLDR: Stand on a tall hill or a mountain, shoot while the rifle is exactly horizontal. Use a top-tier extremely high velocity round, custom-loaded with extra powder, with a high caliber sniper rifle. Anything less would not work.

  I know you need oxygen for the ignition

You don't. Gun powder contains its own oxygen. You can fire guns in a vacuum just fine. Primers also go off just fine in a vacuum.

It should be noted that the important number here, 1.7 km/s, is the orbital velocity near ground level on the Moon. It must not be confused with the Moon's escape velocity (the speed at which an object breaks away from orbit and keeps going away into space forever, never to return), which is higher.


How was the core temperature of the Sun estimated? It was estimated that the heat inside the core of the Sun inside around 15 000 000 °C - this value is extremely enormous. How did scientists estimate this value?

Hydrodynamic models of the Sun allow one method of estimating its internal properties. To do this, the Mass, radius, surface temperature, and total luminosity (radiative energy emitted)/s of the Sun must be known (determined observationally). Making several assumptions, e.g., that the Sun behaves as a fluid and that local thermodynamic equilibrium applies, the stellar equations of state can be used. Numerical methods are applied to these equations to determine the internal properties of the Sun, such as its central temperature.

A great example for how to work this problem your self can be found in the undergraduate text, 'An Introduction to Modern Astrophysics' by Carroll and Ostlie (Section 10.5). The FORTRAN code to run your own stellar model is included in Appendix H.

A comprehensive review paper on how stars of different masses evolve internally (e.g., with respect to T, P, etc.) that is worth reading is:
http://adsabs.harvard.edu/abs/1967ARA%26A...5..571I

A very interesting historical overview of the development of the Standard Solar Model:
http://arxiv.org/abs/astro-ph/0209080

This (admittedly dry) paper gives you a good idea of how well the 'standard' solar models estimate the internal properties of the Sun using helioseismology and neutrino measurements to help tie down their boundary conditions:
http://adsabs.harvard.edu/abs/1997PhRvL..78..171B
The answer is that they match incredibly well (>0.2% error)

These were the least technical (but still academically published) references I could find.

Here is a comprehensive page on the state-of-the-art in solar modelling and measuring the internal Sun using Helioseismology:
http://www.sns.ias.edu/~jnb/Papers/Preprints/solarmodels.html
(highly technical)


Did the scientists really observed gold and platinum being created from kilonova? I read that after LIGO and Virgo interferometers detected gravitational waves signal from a kilonova, many ground based telescopes quickly turns their mirrors towards the spot and "sees" gold and platinum ,etc other heavy elements over the next few days. Is it really possible to directly observe heavy elements being formed consider that the event occurs at a distance of around 130 millions light year away?

No they didn't. The deduction that (very) heavy elements are being produced is somewhat indirect and involves creating a theoretical model of the expanding "fireball" and comparing that model against measurements of brightness vs time at a range of wavelengths and against spectra (mostly in the infrared) as the afterglow faded.

Normally, when we view a stellar spectrum, one can see individual absorption lines that correspond to particular, identifiable elements. In the afterglow of this explosion that is not possible. It is moving so fast that the absorption features that were seen are smeared out by the Doppler effect and the atomic physics required to predict the strength of absorption is too uncertain to make accurate identifications of particular elements.

What is likely though is that the heavy elements, known as lanthanides are providing the large opacity in the "photosphere" of the fireball and are responsible for the rapid fading of the visible signal, the rather more lengthy decline in the infrared and the appearance of extremely broad absorption features that develop in the infrared. The heating of the ejecta by the radioactive decay of very neutron-rich nuclei is also a vital component of explaining the development of the afterglow.

See for example Pian et al. (2017) and Chornock et al. (2017).


Why is Rosetta's Comet so "warm"? Rosetta Comet AKA 67P/Churyumov–Gerasimenko has the surface temperature range of -43 to -93 degree Celsius.

Now, if we compare that with the Mercury (I chose Mercury cause it's closest to the Sun and has the thinnest atmosphere), the minimum surface temperature is -170 degrees Celsius. A little farther from that, on Earth's Moon, while at night, the temperature can drop up to -233 degree Celsius.

Rosetta's Comet is farther from the Sun as compared to both objects (Perihelion 1.2432 AU, Aphelion 5.6829 AU); however, a lot warmer. Why is that?

67P/Churyumov–Gerasimenko is actually not really warm when compared to these bodies.
You quoted figures for the minimum temperatures for Mercury and the Moon. Those are accurate but misleading. Both are for nighttime temperatures, taken from locations on the bodies that are facing away from the Sun. Daytime temperatures can skyrocket, to 427 degrees Celsius during the day on Mercury. The Moon can also get very hot during the day, although nowhere near as hot as on Mercury,
Looking at it like this, 67P/Churyumov–Gerasimenko's surface temperatures are pretty moderate, and nowhere near extreme. Rubin et al. (2015) suggest that it formed in extremely cold conditions - albeit ones not too extreme, for a comet.
The ESA has something related to say on the matter:

At these distances, the comet covered only a few pixels in the field of view and so it was not possible to determine the temperatures of individual features. But, using the sensor to collect infrared light emitted by the whole comet, scientists determined that its average surface temperature is about –70° Celsius.
The comet was roughly 555 million km from the Sun at the time — more than three times further away than Earth, meaning that sunlight is only about a tenth as bright.
Although –70° C may seem rather cold, it is some 20–30° C warmer than predicted for a comet at that distance covered exclusively in ice.

That said, evidence from the mission has made scientists think that an surface with a lot of dust may be the norm for comets.
67P/Churyumov–Gerasimenko isn't warmer or colder than the other bodies at this distance from the Sun. It's simply milder.

I would wager that the reason Mercury and the Moon can get so cold is because they don't really have atmospheres. Over the course of a day, Earth's atmosphere heats up a bit, and over night, it retains that heat, meaning that the surface can stay relatively warm. Mercury's atmosphere is extremely tenuous. The same goes for the Moon.
This might not be an issue, but one day on Mercury lasts 59 Earth days. One day on the Moon lasts for about one Earth month. This means that while there's time to trap some heat, there's also the same amount of time to lose it, and without an atmosphere, the heat gained is soon be lost.
67P/Churyumov–Gerasimenko also doesn't have an atmosphere to speak of, but it rotates quickly - rotating once every ~12.5 hours. This isn't necessarily perpendicular to the orbital plane, so not every part of it will face the Sun during a rotation, but it does mean that there's not as much time to lose heat.


How far can you see in solar plasma at just under 1 solar radius? I was trying to compose an answer to this question here, about touching the surface of a star, and I was going to mention that the popular depiction of the sun's surface as 'basically lava' (as depicted in Sunshine, for example) is wrong.

The usual definition of a star's radius is where the optical depth is 2/3. At this radius, the density of the sun is about 1/10th of air at sea level and can hardly be described as solid or liquid.

I wanted to get a sense of what this diffuse plasma would look like if you were in it and somehow impervious to the heat and radiation. Would it appear as a thin, glowing fog, through which you could still discern distant objects? Or would it appear as opaque as, say, a bonfire?

Rephrasing in a little more technical way, What is the extinction coefficient of solar plasma at ~1 Solar radius in the visible spectrum?
What is the path length at which all light is effectively attenuated?

(Since it varies by wavelength, I mean specifically the region of the visible spectrum where the plasma is at its most transparent).

If you have a look at the top-left panel of Fig.11 in these lecture notes by Rob Rutten, you will see that the continuum opacity at optical wavelengths at the photosphere is about $10^{-6.7}$ cm$^{-1}$.

The inverse of this is the optical depth. You can see stuff through about 2-3 optical depths, so your "horizon", looking horizontally in the solar atmosphere is about 100 km.

However, the same plot shows that, because of the temperature and density variations with height, the photosphere becomes an order of magnitude more or less opaque at only 50km below or above the "photosphere" respectively.

What this means is that you can see about 100 km horizontally, but less than this (guesstimate 30km) looking downwards and more than this (guesstimate infinity - that is how the photosphere is defined!) looking upwards.

These numbers would all be revised downwards at the wavelengths of prominent absorption lines.

If you could look all around you, you would see a very asymmetric brightness/colour distribution. Overhead would be dimmer and cooler (redder); underneath, much brighter and hotter (more or less the same colour we see the photosphere).


Peculiar Amplitude of Earth's Orbit in Z Axis I'm currently building an N-body simulator for my masters project and I'm using JPL's Horizons system to compare against.

I started with just a couple orbits and everything seemed to be working nicely (After fixing many mistakes implementing a Fourth Order Integrator) but after extending to 10+ orbits I notice that in the Z-axis the Earth's amplitude increases linearly, according to JPL, but my simulation has a constant amplitude. JPL's result is shown in red in the bottom left:

What is causing this? Currently I only have the Sun, Earth, Moon and Jupiter in the system, but I didn't even realise the amplitude increased like this? What am I missing or got wrong?

(Also, the graph only shows 3 simulated orbits, and 80 JPL orbits)

I can reproduce your JPL Z coordinate plot if I use Earth's heliocentric position relative to the J2000 ecliptic.
Naturally the amplitude has a minimum around the year 2000:

Relative to the ecliptic of date, which accounts for precession, the heliocentric Z coordinate of the Earth-Moon barycenter is much smaller:

The IAU 2006 precession model has two components, detailed in Capitaine et al. 2003:

precession of the ecliptic, a 47"/century change in the plane of Earth's orbit
precession of the equator, a 1.4°/century change in the plane and axis of Earth's rotation

Fifty years of ecliptic precession amount to 23.5" or 0.000114 radian, and the first plot is consistent with that.
The classical precession of the equinoxes (the intersection of the two planes) is a combination of both components.

Williams 1994 examines Earth's equatorial precession (dominated by the Sun and Moon) and finds that Venus's effect is larger than Jupiter's.
When looking at fluctuations in Earth's year length, I noticed a cycle similar to Venus's synodic period.
I expect that adding Venus to your system would reproduce some ecliptic precession.


How do you find the altitude of the Sun if you are on the Moon? The question is:

  If you stand on the equator of the Moon, at midday your position occurs when the Moon is located at the line of the node. What is the altitude of the sun at that moment?

Information: This is when the Moon is located at the line of node (line of node is a line that occurs from crossing between the ecliptic plane and the Moon's orbital plane, which makes an angle of approximate 5 degrees). The equatorial plane of the Sun makes an angle of 1 degrees, 32 arc minutes and 40 arc seconds.

The solution is 88 degrees, 27 arc minutes and 20 arc seconds through 90 degrees.

But I don't understand why the answer isn't 1 degree, 32 arc minutes and 20 arc seconds, because the ecliptic plane where the sun is located is above the horizon or celestial equator. equal to its altitude.

@DavidHammen is essentially correct. According to stellarium, the Sun gets as high as 89 58' 48":


How many GAIA Data Release 2 files are there? I am using this index page to download GDR2 data files.

http://cdn.gea.esac.esa.int/Gaia/gdr2/gaia_source/csv/

But it keeps giving a different number of links. Sometimes it shows 16,000 files, sometimes 31,000 and the max I have seen is 61,237. What is the correct number I should download? GAIA GDR2 documentation does not list the total number of files available.

Interestingly, if the index lists less than 60k files, it still has the closing /html tag so it would appear I have a complete index file, but the contents are not complete.

There are 61,234 data files in the directory. There are three additional (non-data) files:

MD5SUM.txt
_citation.txt
_disclaimer.txt

If you include those, the total number is 61,237.


Are high-speed galactic collisions survivable? This is a hypothetical question, but it really is meant as a vivid way of asking about the affects of high-speed galactic collisions. The November 27 APOD showed the Cheshire Cat galaxy group, and the caption says that the "eyes" are two large elliptical galaxies, closing on one another at over 1000 km/s, and bathed in a glow of X-rays from heating of the associated interstellar media.

How intense is this glow? I suspect it is very faint, but I don't know how to quantify it. Would it hinder X-ray astronomy for residents of the galaxies? Would it be strong enough to affect life? Does anybody have quantitative information and know how to interpret it? (I do realize that elliptical galaxies are thought to be made up of old, low-metallicity stars that aren't good candidates for habitable planets, but let's ignore that.)

Interesting question.   I did a bit of reading and gave it some thought and and I can sort of give an answer, though I invite corrections and input from anyone more knowledgeable than me.

First, those 2 galaxies are enormous.  I didn't see specific sizes listed, but per this paper:

  1.2-1.5 x 10^14 solar mass fossil group

That's about 100 Andromeda galaxies.  Also, since what we're seeing is 4.6 billion years old, it's reasonable to assume that those 2 galaxies probably have more free dust and gas and, being larger, the gravitational collision is happening at higher speed.

That's really the question with how much background x-ray radiation you get when 2 galaxies collide.   how much free gas and dust and how fast the galaxies move into each other.   My guess is that being larger, and younger, those 2 galaxies are glowing much more brightly than the Milky-way and Andromeda will 4 billion years from now.

As to how intense the glow is, my hunch agrees with yours, it's probably not very bright if you're there.   The faint glow and the 2 bright eyes you see in the picture are long exposures.   This site (scroll to the bottom) says it's 19 hours and 30 minutes of exposure.  Andromeda, for example is barely visible to the naked eye, but when photographed it's made many times brighter.   I'm not smart enough to say for 100% sure, but my guess is that the faint purple glow would be barely visible, if visible at all, assuming it even reached into the visible spectrum, which X-ray's don't.  Chandra might be setting the color to a visible spectrum in the picture.

How dangerous is it?   Again, I'm guessing, but I think in the Milky-way-Andromeda collision, we'd probably get more UV from the sun than we would from high speed gas and dust collisions, but that's just a guess.   It would be kind of fun if the entire night sky glowed a little bit though.

As far as the Milky-way-Andromeda collision, Wikipedia says that there won't be much free gas in the collision, here, so it's a safe bet there won't be any visible glow in the sky and not even that much of an acceleration of new star formation, though there should be some.   The Milky-way currently forms about 7 new stars every year, and that I imagine will gradually slow down over the next 4 billion years, then speed up significantly as the collision gets underway, but from the point of view of Earth, I don't think there's any guarantee we'd get a close look at any new star formations.

Here's a fun article/interview on the collision between the Milky Way and Andromeda, though I think Roeland van der Marel might be exaggerating the new star formation a bit if we're to believe Wikipedia where it says there won't be much free gas.

As for the speed of our collision, Andromeda is currently heading towards us (or, we're heading towards it, whichever you prefer) at about 110 km/s source.  and it's currently about 2.5 million light years away.  Source.

Covering 2.5 million light years (about 24 million trillion km) in 4 billion years (126 million billion seconds), works out to an average speed of 190 km/s, so we can roughly estimate that the speed of impact between our two galaxies will reach a peak somewhere around 270 or so km/s.   But (see video you also have to take into account the rotation speed and it appears (video above) the rotations will be going in opposite directions when the galaxies collide.    We orbit the Milky way at about 250 km/s and Andromeda, being larger, probably a bit faster orbit speed.   Adding the 2 velocities together, in our neck of the woods, we might see some relative velocity and some gas & dust collisions at (guessing here, cause I don't know if the rotations will slow down as the galaxies approach), but lets say 500 to 600 km/s or so.   For any close bright stars from Andromeda, that could be fast enough for visible changes to a few near-by Andromeda stars in constellations over a single human lifetime.

Perhaps, we'll pass through an oort cloud of another star every few thousand years or so, perhaps even the occasional kuiper belt equivalent, every ten or 50 million years - er, maybe.   We could see some very impressive meteor showers and perhaps an occasional, slightly more frequent, dinosaur killing level comet or meteor impact - but I'm just speculating.  There might be lots of interesting things to see between 4 and 5 billion years from now.

When stars form they can be (per this website) 100 to 200 times brighter than they are during their main sequence, so if our solar-system finds itself nearish to a colliding gas cloud and some new star formation, that could be interesting.  Likewise if we pass near the core of Andromeda we might, for a while, have a very bright night sky.


Why are helium resonance lines called "resonance lines"? Examples of the use of the term:

Formation of the helium extreme-UV resonance lines
On the Formation of the Resonance Lines of Helium in the Sun (unpaywalled)
Formation of the helium EUV resonance lines.

When I hear of a resonant state I think of a free particle incident on a potential where it spends a lot of time then leaves, but the electrons responsible for these lines (HeI 584 Å, HeII 304 Å) start in bound states below their binding energy and are just transitions from one bound state to another.

Since most emission lines from atomic transitions are not called resonance lines:

Why are these lines different?
What is particularly resonant about them?

It appears to be a conventional label that is applied to transitions between the ground state and another energy level (some definitions specify the first excited level) of an atom and is used in all the physical sciences, not just astrophysics.
e.g. He I (58.4 nm) is a transition from $^1$P to the $^1$S ground state.
In fact all atomic/ionic transitions can be considered resonant phenomena, but the term "resonance line" is applied only to these particular types of transition, perhaps because they are usually the strongest lines in the spectrum from that species, since in most cases, the ground state is the most populated. So there is nothing "particularly resonant" about them
A "resonance line" is:

A spectral line caused by an electron jumping between the ground state and the first energy level in an atom or ion. It is the longest-wavelength line produced by a jump to or from the ground state. Because the majority of electrons are in the ground state in many astrophysical environments, and because the energy required to reach the first level is the least needed for any transition, resonance lines are the strongest lines in the spectrum for any given atom or ion.

https://www.oxfordreference.com/view/10.1093/oi/authority.20110803100415843
And here it is used in that context in astrophysics: (Morton 2013)

The tabulation emphasizes resonance lines, i.e., lines whose lower level is the ground state.

Or in Chemistry: From https://goldbook.iupac.org/terms/view/R05341

The radiative decay of an excitation level may proceed to the neutral ground state and would thus occur at the same energy as the corresponding line in the absorption spectrum. Such a line is called a resonance line and the process is called resonance emission.

Or from the Basic Atomic Spectroscopic Handbook:(https://www.nist.gov/pml/basic-atomic-spectroscopic-data-handbook)

The strongest persistent lines usually include one or more resonance lines, i.e., transitions to the ground level or term.


Question about Telescopes , observations and explanations of the observations Though I am active on Physics, I am new to Astronomy. I am an amateur astronomer, actually even worse than amateur. I have just started it.
I have a 60 mm objective 700mm focal length refractor which I mainly use with a 20 mm eyepiece.

I have a couple of questions regarding the telescope, observations and explanations behind observations. Since they are all interlinked I decided to ask them together.

1) So, firstly the basics. Regarding the resolution power of a telescope, we know from the Rayleigh criterion that diffraction does happen at the objective. But Diffraction happens when light passes through a narrow slit or bends around a corner (whose dimensions are of the order of the wavelength of light). But there is no slit and probably no bending of light. How come then do we assume a diffraction pattern to have formed when considering the derivation of the resolution of a telescope?

2) How are exit pupil and entry pupil different from the diameter of the eyepiece and objective?

3) Why on increasing the magnification does the brightness of the object reduce (the physics behind it)? I mean that if I use 4mm eyepiece then, first of all, it becomes very difficult to find the object and secondly its brightness gets reduced when compared to the brightness that I had got using a 20mm eyepiece?

4) Secondly why do the Barlow lenses reduce the brightness a lot. And why on using  Barlow lenses the field of view get smaller and the object passes twice as quickly as without the Barlow?

5) Now the biggest question. How is it possible that every time I point my telescope towards a visible planet or a star (I know it's a star because of the sky map), I see almost the same size, when stars are visible only as small dots. The stars appear to be bluish and the planets which I have seen (Mars and Jupiter) appear reddish. That's ok but a star can not be visible as a sphere. Stars appear as dots, don't they?

6)How can I make the telescope give the best results i.e which eyepiece should I use and whether I should use Barlow lenses or not?

I'll attempt to answer as many of these questions, to the best of my abilities, as I can at this late hour.

1) Lenses have an index of refraction that is different than vacuum, and air, and they are curved. ie: corners when light is concerned.

2) The entrance and exit pupils are the pupil size required to pass an "extreme" ray through an optic. Meaning that if the light ray from the optical axis was not refracted at all as it passed the optical plane, it would still pass within the pupil (entrance) and if reflected perfectly (exit). The diameters of the objective and the eyepiece are subjective, meaning that you can have, within reason, any size objective and use virtually any eyepiece. The restrictions to this would only be focal lengths.

3) You are looking through a thicker lens and thereby reducing the light incident on your eye. Think about the shape of the lenses, look at them and it might become apparent just how much more curvature the 4mm lens has than the 20mm lens. This added curvature comes at the cost of a higher angle of incident that a light ray has to take, and thus more optic to pass through.

4) Barlow lenses reduce the brightness a lot because you are adding additional optics. With each new addition you decrease the amount of light by a factor relative to the thickness and clarity of the optics. The Barlow lens is increasing the magnification and thereby reducing the field of view.

5) With your specific telescope (of which I have one with the exact same size/focal length) you should be able to see Jupiter as a disk, Mars too should be more than a "dot." Stars are point sources with angular diameters much too small for you to be concerned given the size of your telescope. Also, those planets appear reddish because they are-ish. Many stars will still appear colors other than blue if you can find them.

6) This is up to your discretion and desired results when observing. Please read your owners manual, or check out the telescope mfr website for recommended setups.


Does lightning on Venus affect the chemical composition of the clouds? It is established that lightning occurs on Venus due to the presence of sulfuric acid clouds. Does this constant electric discharges and possible electrolysis affects the composition of the clouds on Venus for instance, sulfuric acid dissociating into ions like hydrogen ions, sulfate/bisulfate ions or decomposing into sulfur oxides or changing to other sulfur oxoacids?
There was a question on space.SE that discussed if peroxydisulfuric acid (H2S2O8) created due to electrolysis of concentrated sulfuric acid (from the lightning) can be used as a propellant fuel. If the species is indeed formed in the atmosphere, how abundant is it? How much of this species is present in the atmosphere?
Does lightning on Venus change/affect the chemical composition of the clouds? Does it affect the sulfur cycle of Venus in any way (so that it could create more exotic sulfur species)?

Observations by many spacecraft that have visited Venus over the last
40 years appear to confirm the presence of lightning storms in the
Venus atmosphere. Recent observations by Venus Express indicate that
lightning frequency and power is similar to that on Earth. While
storms are occurring, energy deposition by lightning into Venus
atmospheric constituents will immediately dissociate molecules into
atoms, ions and plasma from the high temperatures in the lightning
column (>30,000 K) and the associated shock waves and heating, after
which these atom and ion fragments of C,O,S,N,H-containing molecules
will recombine during cooldown to form new sets of molecules. Spark
and discharge experiments suggest that lightning
effects on the main atmospheric molecules CO2, N2, SO2, H2SO4 and H2O
will yield carbon oxides and suboxides (COm, CnOm), sulfur oxides
(SnO, SnOm), oxygen (O2), elemental sulfur (Sn), nitrogen oxides (NO,
N2O, NO2), sulfuric acid clusters (HnSmOx−.aHnSmOx e.g. HSO4−.mH2SO4),
polysulfur oxides, carbon soot and other exotic species. While the
amounts generated in lightning storms would be much less than that
derived from photochemistry, during storms these species can build up
in a small area and so their local concentrations may increase
significantly. [...]

Source: M.L. Delitsky, K.H. Baines, Storms on Venus: Lightning-induced chemistry and predicted products, Planetary and Space Science, Volumes 113–114, 2015,Pages 184-192, 10.1016/j.pss.2014.12.005


Will the Sun's solar wind push a permanent magnet out of our solar system? If you were place a strong permanent magnet into the solar wind, like a 1 inch cube neodymium magnet, would the charged particles of the solar wind colliding with the magnet's magnetic field push the magnet out of our solar system, that being if it does not encounter any planets or objects in its path?

No. The magnetic field has two poles, the force on the two pole of the magnet is equal and opposite. This is why a compass needle will align in the North South direction, but is not pulled towards the North or the South.

Two magnets will attract or each other because the North pole of one is nearer the South pole of the other. You can get a net force on a magnet when the field-lines are getting bunched together. But the magnetic field in the plasma is not like this. To a first approximation they are parallel to each other and perpendicular to the equator of the sun. In fact the magnetic field is complex as it is carried with the flow of solar wind.

There would be no net force on the magnet, but it may tend to align with the magnetic field. This might cause a turning force, but no acceleration.

There would be some force due to solar wind particles impinging on the object, but since there is only something like 5 atoms per cm3, the force isn't very big, and it has nothing to do with the object being a magnet.


How do black holes evaporate? Will black holes evaporate, if they evaporate? When and why do they evaporate? What are the conditions for evaporation? Under what principle do they evaporate?

The evaporation occurs through Hawking radiation. This is a very slow and low energy process. So low that the cosmic microwave background radiation, which is just a few degrees above absolute zero, pours far more energy into the black hole than Hawking radiation takes away. So in principle a black hole cannot evaporate.

With the exception of conjectured atomic sized black holes, that is, as Hawking radiation would be more pronounced then.

Edit: See this answer on the physics SE: https://physics.stackexchange.com/a/26607/55483
The black hole would need to be less massive than the moon to radiate more energy than it absorbs. Bigger than atomic size, as I suggested.


How can I get/generate/extract PSF image from galaxy .fits image? I want to model an image of a galaxy from Hubble.
For this, I need the PSF image as well.
I already tried to use TinyTim, but I don't have half of the parameters and cant be sure about the quality of output.
Is there any way for me to get it from the .fits image of a galaxy?

To answer the question: Yes, there is a way. Use a bright but unsaturated star. That might already be good enough, or you might want to do any number of corrections to this PSF model.
A better way is to figure out the parameters to use with TinyTim, and generate an approximately right PSF from those. Most parameters you know, or they are easy to look up, such as camera, chip, filter. For the pixel position, you need to look up the original exposures that went in to your (presumably) drizzled image; the object will be more or less in the same place in all of them, so you can use the average position.
With these, you can generate an overresolved PSF image. If your (presumably) drizzled image is north-aligned, you will need to rotate the PSF accordingly; ideally using drizzling, but a simple image rotation might work well enough if your PSF resolution is much higher than the HST resolution. The final result is a PSF that might not be useful for precision photometry, but should be good enough for modelling a galaxy. In any case, compare the PSF to the image of a star to make you it is roughly right (east-of-north angles in HST images have a way of being wrong).


What is the ugriz magnitude system? While studying photometric redshifts, I came across the ugriz (or u'g'r'i'z) system for classifying magnitudes of galaxies, but I didn't find much information on the internet about how it works.

Can somebody explain that? Do the letters "ugriz" have any special meaning, or is its just a nomenclature?

"ugriz" is short for U-band, G-band, R-band, I-band, and Z-band, detailed on the Wikipedia article on photometric systems:
$$\begin{array}{|c|c|}
\hline \text{Band}&\text{Effective Wavelength Midpoint }(\lambda_{\text{eff}})\\
\hline \text{U} & \text{365 nm}\\
\hline \text{G} & \sim\text{475 nm}\\
\hline \text{R} & \text{658 nm}\\
\hline \text{I} & \text{806 nm}\\
\hline \text{Z} & \text{900 nm}\\
\hline
\end{array}$$
This $ugriz$ system is slightly different from the $u'g'r'i'z'$ system; conversions can be found here. The starting point is the same (note that $u(\text{2.5m})=u'$), but there are slight vertical and horizontal shifts.

The differences between $ugriz$ and $u'g'r'i'z'$ are much smaller than the differences between $u'g'r'i'z'$ and the $UBV(RI)_C$ system, which was replaced by them both. This older system of filters had large overlaps and did not cover as broad a range. See Fig. 1 from this thesis for a comparison:


Moon's orbit around the Sun The Earth revolves around the Sun and the Moon revolves around the Earth.
Out of curiosity I started thinking about the orbit of the Moon around the Sun and expected (assumed) it to be as follows:

But on Wikipedia and some other sites I found out that the orbit is actually like this:

I have 3 questions:

What is the reason for this difference between assumed and actual path variation?
Has this path been like this since the formation of the Moon?
Do natural satellites of other planets also follow the same orbit around the Sun?

After further searching I found out a better, easier explanation regarding the orbit path on YouTube for those interested, be sure to check it out.

Not an answer, but I thought this was a good slice of a picture of the Moon's orbit around the sun.

Source:  http://www.wired.com/2012/12/does-the-moon-orbit-the-sun-or-the-earth/


How was water formed? We know that a planet is formed from the nebula of gas and dust. Gravity pulls the particles together in a core. So we all know that the gas cloud is a cloud of hydrogen.

But how did water form? From where did oxygen come? How did it combine to make $\text{H}_2\text{O}$?

Stars, especially larger stars, are hot enough at their core to produce significant amounts of heavier elements: Carbon, Nitrogen and Oxygen. When the stars die, some of these elements are returned to the interstellar medium.

The nebula from which stars form contains not only Hydrogen, but also all the other elements found on Earth. The nebula contains the Carbon that is the basis of life, the Nitrogen that forms the atmosphere, the Silicon that makes rocks, and the oxygen that reacts with hydrogen to form water.

Most of the water on Earth was actually brought to the Earth on asteroids after the Earth had formed, see How did water get on Earth


Absolute/apparent magnitude and distance for HIP31978 inconsistent? EDIT: I am still following up, but here's the latest response from Wolfram:

  Paclet servers are updated and optimized for each version of
  Mathematica, so I suspect you may be working on an older version which
  we no longer update. Many times we get reports that data from the
  paclet data for version 9 was incorrect, only to find that it has
  already been fixed for version 10.

According to Mathematica and several other sources, the
known information about HIP31978 (also known as "S
Monocerotis" and "15 Monocerotis") includes:

The star system is about 101.06 light years from our own.
The star's absolute magnitude is -2.79 (some sources say even lower).
The star's visual magnitude from our solar system is 4.66.

The star is slightly variable, but not enough to explain
the following discrepancy:

If the star's magnitude at 32.6 light years is -2.79
(definition of absolute magnitude), it would be about 9.6
times fainter at 101.06 light years (it's actual distance
from us). However, that's only about 2.5 magnitudes
fainter, nowhere near enough to bring it down to magnitude
4.66.

What's happening here, and is this a one-off special case,
or does this happen a lot?

For reference, my calculation of the apparent magnitude:

-2.79 - Log[100, (32.6/101.06)^2]*5 == -0.333192

The absolute magnitude you quote appears to correspond to the Hipparcos parallax distance estimate, which is $\sim 1000 \mbox{ ly}$, not the $\sim 100 \mbox{ ly}$ distance you have used.


what is the resolution of the magnitude that is distinguishable to the naked eye? Or is it possible to spot the changing brightness (magnitude) of a planet with a naked eye

The context of the question is this: If I'm not mistaken Ptolemy was able to say (using observation alone) that the retrograde motion of a planet happen to be when it was closest to earth (that behavior was consistent with his epicycles model).

So I have my doubt with respect to the accuracy that this kind of observation could be done at all.

While the eye is terrible at determining apparent magnitude or brightness of stars due to our adaptive iris, it's perfectly capable of discerning relative brightness between two objects in the same visual field.

Also, the ancient Greeks could discern six magnitudes of star brightness by realizing that brighter stars look bigger in the night sky.  The word 'magnitude' means 'bigness' or 'size'.

Magnitude - Wikipedia

This wasn't the most accurate system, but it was good enough to be able to discern about six different magnitudes. The system dates back about 2,000 years, to either Hipparchus or Ptolemy.   So there's no question that ancient greeks had at least a rough categorization of stars by brightness.

Now, could they tell when a planet was closest?  Easily.  The difference in magnitude between closest and farthest distance from the Earth for most of the planets Ptolemy could see was very large.  For example, Mercury at its brightest as seen from Earth has an average apparent magnitude of -1.89, and at its faintest, +5.93.  A change in magnitude of that amount would be easily discernible by the human eye. Venus differs by about two magnitudes.  Mars almost five magnitudes.  Jupiter about 1.3.  Saturn is the only one with a difference of less than one magnitude, making it uncertain if the ancient greeks could see that difference.

Calculating Apparent Magnitudes of Planets

If the ancients couldn't determine the absolute magnitudes of the planets, they could certainly compare them to the magnitudes of known stars and determine if they were brighter or dimmer.  Given the huge magnitude swings of the inner planets, determining when they were closest would have been relatively easy, within some margin of error.


Up to 384 minor planets (including Pluto) qualify as planets? In 2006, the International Astronomical Union redefined the definition of planet in order to exclude Pluto, Eris, and several other objects whose category was disputed. This new definition of a planet featured 3 criteria:

A planet must be massive enough to achieve hydrostatic equilibrum and
become ellipsoidal.
A planet must orbit a star or brown dwarf (with the exception of planemo).
A planet must be massive enough to clear any nearby objects.

Pluto and the other objects apparently failed to meet the 3rd criteria, as they had not cleared the nearby Kuiper belt objects, so they were reclassified under the term "dwarf planets".

But what if Pluto actually has met the third criteria? It's probably well-known by now that Pluto has 5 moons. The largest moon Charon is over half the size of  Pluto, and is frequently classed as a dwarf planet in its own right. The other 4 minor moons Nix, Styx, Hydra, and Kerberos, are much smaller then Charon, and are similar to the many asteroid-like moons found around gas planets.

The leading theory for how these smaller moons formed is that they are leftover debris from a collision between Pluto and another object, but it's just as likely that they were Kuiper belt objects that were captured by Pluto (just like the moons of Jupiter and Saturn). If they were captured, this means that Pluto has technically cleared 4 objects from its neighborhood, and so it meets the 3rd criteria.

The same logic can be applied to Eris, 2007 OR10, and the 400 other dwarf planets known to host moons. So should we have 392 planets, keep the current definition, or scrap or redefine the term "planet" entirely?

That actual IAU Resolution B5 adopted at the IAU General Assembly in 2006 states:

(1) A planet is a celestial body that:
(a) is in orbit around the Sun,
(b) has sufficient mass for its self-gravity to overcome rigid body forces so that it   assumes a hydrostatic equilibrium (nearly round) shape, and
(c) has cleared the neighbourhood around its orbit.

The last part being the neighbourhood around its (whole) orbit is the important part; it's not enough for Pluto to dominate its family of moons (and this last part is debatable as Pluto and Charon orbit a common barycenter (center of mass) which is between Pluto and Charon since Charon is so massive, compared to Pluto).
Since Pluto orbits in the Trans-Neptunian region with a similar orbit to other bodies, some of which are similar or larger in size and mass (Eris), and cross the orbit of Neptune, it can't really be said have to "cleared the neighbourhood around its orbit". Indeed Pluto is the prototype of the Plutinos; object which have a 3:2 mean motion resonance with Neptune (orbiting twice for every three orbits Neptune makes).


How to find a deep sky object 1° NE from a given star I am an amateur so sorry for any misunderstanding.

I am trying to find a deep-sky object located 1° north east from the δ Cas star. I am using the Stellarium program to do this.

I have toggled on the equatorial grid and Cardinal points are displayed as well, however, all of the cardinal points appear to be on circumference and I cannot really tell which direction is the wanted object at.

What am I doing wrong?

Usually when the direction to an object is given relative to a star, the directions are related to equatorial coordinates (the coordinates on the sky). They are not related to the compass or cardinal points. It is best to avoid directions relative to the horizon/compass/cardinal points since those depend on the time and location.

In your example, 1 degree northeast of $\delta$ Cas means this direction:

north is toward Polaris along a line of right ascension, so you would go in the direction of increasing declination from $\delta$ Cas in the direction of Polaris.
east is along a line of declination in the direction of increasing right ascension.

When looking at the sky, and if north (the direction toward Polaris) is thought of as the 12 o'clock position, the direction north east would be approximately 10:30. (Likewise, east would be the 9 o'clock position, south would be the 6 o'clock position, and so on.)

Here is a view of Cassiopeia for an arbitrary time and date with the directions shown. Are you trying to locate M103?


When could the Roche limit be larger than the Hill sphere? This answer to Are Saturn's rings stable? begins with:

  Most of Saturn's rings are inside it's Roche limit, which means they will never clump together. Tidal forces prevent this from happening.

I always confuse Roche limit and Hill sphere, so I thought I'd ask a question about both.

Question: Under what conditions could a solar system body orbiting the Sun have a Roche limit larger than its Hill sphere?

The Hill sphere is approximately given by

$$R_\mathrm{Hill} \approx a_\mathrm{p} \left(\frac{M_\mathrm{p}}{3M_\ast}\right)^\frac{1}{3}$$

Where $a_\mathrm{p}$ is the radius of the planet's orbit, and $M_\mathrm{p}$ and $M_\ast$ are the masses of the planet and the star respectively. This is an approximation to the size of the Roche lobe around the secondary.

The Roche limit (not to be confused with the Roche lobe) is the limit at which tidal forces will disrupt an object held together by its own gravity. Roche derived the following formula:

$$R_\mathrm{Roche} \approx 2.44r_\mathrm{p} \left(\frac{\rho_\mathrm{p}}{\rho_\mathrm{s}}\right)^\frac{1}{3}$$

Where $r_\mathrm{p}$ is the radius of the planet, and $\rho_\mathrm{p}$ and $\rho_\mathrm{s}$ are the densities of the planet and the satellite respectively. Note that there are various different formulae for the Roche limit depending on the different assumptions being made, see the Wikipedia Roche limit article for details. For a spherical planet, this can be rewritten in terms of the planet's mass:

$$R_\mathrm{Roche}
\approx 2.44 \left(\frac{3M_\mathrm{p}}{4 \pi \rho_\mathrm{s}}\right)^\frac{1}{3}
\approx 1.51 \left(\frac{M_\mathrm{p}}{\rho_\mathrm{s}}\right)^\frac{1}{3}$$

So the condition you're interested in becomes:

$$1.51 \left(\frac{M_\mathrm{p}}{\rho_\mathrm{s}}\right)^\frac{1}{3} \gtrsim a_\mathrm{p} \left(\frac{M_\mathrm{p}}{3M_\ast}\right)^\frac{1}{3}$$

Cancelling and rearranging gives:

$$a_\mathrm{p} \lesssim 2.18 \left(\frac{M_\ast}{\rho_\mathrm{s}}\right)^\frac{1}{3}$$

For a satellite density of 3300 kg/m3 (similar to the Moon) with a host star the mass of the Sun, this corresponds to a planetary orbit of about 2.6 solar radii.

Needless to say, this is an extremely close star–planet separation that invalidates a lot of the approximations used to compute that limit. For example, the planet will be close to or within its Roche limit with respect to the star. If it is not itself being disrupted, it will likely be significantly non-spherical.


What does it mean when we say that information is potentially recoverable from a black hole My understanding is that if the surface of a black hole is holographic, then information is potentially emitted from a black hole, say due to Hawking radiation or some other means.

What can say about this information?  Is it physically possible and/or feasible that this type of emission would embed information about what went into it?  Also, how does this measure up against the idea that the world lines inside the event horizon are completely disconnected from the outside?

I have some knowledge of the information paradox, but I was really hoping someone could  explain it in simple terms or point me to a prior explanation along those lines.

The first thing to understand is that we have no actual knowledge of how information and black holes interact.

What we do have are theories that are well supported by experiment (General Relativity, which predicts Black Holes but which says nothing about information) and Quantum Mechanics (which says a lot about information (in this technical sense) but nothing about BHs.)  We also have a variety of theories, none of which have any substantial experimental support, and all of which are incomplete (which means we don't know if there's actually a viable theory there -- all we have today is bits and pieces that look promising).  One of the goals of all of these theories is to -- somehow -- meld GR and QM into a single coherent theory with greater explanatory power than the two separately.

It is very fair to say that some of the work which has been done to date which tries to link information and BHs is plausible and (at least) does not contradict any experimental evidence we have.  And some of them (like what Hawking used in the prediction of Hawking radiation) are such a careful application of QM in the GR context that most physicists would be surprised if they turned out to be incorrect.

The bottom line is that GR and QM are both amazing theories which yet cannot be completely correct -- they must both be subsumed in some more encompassing theory. (Or at least everyone is convinced of that.  I am too, BTW.)

Due to the mathematical underpinnings of QM and GR, physicists vaguely can see how some generalization of QM might, possibly be able to include GR someday.  But virtually no one can see how GR could be extended to include QM.  So the smart money -- just about all the money, actually -- is on extending QM in some way.

One of the most basic features of QM is that it is linear and preserves information (in a certain well-defined, but technical sense).  If GR could be derived from some extension of QM, we'd expect that that extension of QM preserves information, and that GR also must preserve information.

On the face of it, GR by itself says that information which falls into a BH is lost. Zap. Gone. Never to return.  Therefor finding ways to get GR to preserve information may be an important clue to the ultimate combined theory.  And extensions of GR which preserve information seem more likely to be correct.

So the paradox is that BHs, which are predicted by our marvelously correct theory of gravity, GR, do not conserve information.  But we have reasons to strongly suspect that BH built from the hypothetical gravitational extension of QM would preserve information.  Something's wrong here!

One of the plausible theories that have been developed to resolve this says that Hawking radiation -- which seems very plausible, and which carries energy away from a BH in contradiction of vanilla GR -- might also carry the "lost" information away, eliminating the paradox.

Maybe.

But remember that there isn't any experimental evidence for any of this.  It's all based on smart hard-working, competent scientists trying to use their brains and their intuition to guess what the undiscovered theory which merges GR and QM will predict and to turn it into consistent math.  It's certainly better than a random-assed guess, but it is a far cry from established science.

Keep watching -- I hope that all this work will bear fruit and a real theory with experimental evidence behind it might emerge in our lifetimes.


What is the illuminance of Tau Ceti? As part of my physics project, I investigated the relationship between a light bulb's illuminance and the distance from the measuring device to it.

Illuminance was measured in flux, and distance in metres.

I derived an inverse relationship. First of all, can I check if this is the relationship I was supposed to get? (I understand that there is an inverse square relationship between luminosity and distance, but since illuminance is measured in lumens/sq metre, then I think an inverse relationship is correct...)

Next, using this relationship, I need to extend my project so that I can find distances to stars using this relationship.

As long as 2 stars have a similar mass/size/density, then I can use this relationship to find the distance to the 2nd star as long as the illuminance of the stars are known, and the distance to the 1st star is known.

So I tried to do this with an example - the Sun and Tau Ceti, as they are reasonably similar.

I have values for the Sun's distance/illuminance - $1.5\!\times\!10^8 $km, $1.2\!\times\!10^5$lux.
I now am trying to calculate the distance to Tau Ceti using this method.... But nowhere can I find a value for its illuminance.

Please can someone help me with this? Is there a way to convert from apparent magnitude or luminosity to illuminance?

Thank you

EDIT: The method I want to use from the inverse relationship is:

$$d_2=d_1l_1/l_2$$

where $d_1$ and $l_1$ are the known distance and illuminance of the 1st star, whilst $d_2$ and $l_2$ is that of the second, and $d_2$ is unknown.

You can easily convert flux to luminosity and vice versa using the spherical nature of the propagation of light. The equation you need is:

$$
L=4\pi D^{2}f
$$
Where $f$ will be your flux, and $D$ is the distance to the object, Tau Ceti in this case.

From Pijpers (2003) Tau Ceti's luminosity is $0.52\pm0.03L_{\odot}$. You should have everything you need there. $L_{\odot}$ is of course the luminosity of the Sun.


Can the Sudarsky's gas giant classification be applied to ice giants? Based on the temperature of a gas giant around another star, I have come to understand that it is possible to guestimate its appearance; a classification scheme for gas giants developed by David Sudarky.
I'm wondering if this scheme could plausibly be used to get an idea of what ice giants - Neptune and Uranus like planets - around other stars might look like? If not, are there any resources concerning this topic?

That classification scheme originates from before we had much in the way of observational data on exoplanetary atmospheres. It's not clear that it's a particularly useful description of how hot Jupiters behave: recent models, e.g. Gao et al. (2020) seem to indicate that silicate clouds and hydrocarbon haze are the main factors at play on hot Jupiters, rather than silicates versus alkali metal absorption.
As regards is applicability to ice giants, these planets have substantially higher atmospheric metallicities than gas giants, so it's not clear that these models would apply. For a study of some warm Neptune atmospheres, there's Fortney et al. (2020), who note that the cooling history of the planet is important, and things like tidal heating at depth caused by an eccentric orbit can have significant consequences. Or for a more experimental approach, see He et al. (2020), who indicate that hazes can form in hydrogen-rich atmospheres even in the absence of methane in the initial gas mix. It's a lot easier to find studies on hot and warm planets as these are the ones that are accessible to observations, so they attract more attention: habitable zone terrestrials are an exception to this pattern.
In summary, it's complicated, and things have moved on quite substantially since the Sudarsky et al. papers.


Building a floating, ocean-going giant radio telescope? Given the loss of the Arecibo telescope: Is anybody aware of any efforts or studies towards a floating, ocean-going radio telescope of with 500m diameter and more?
The idea is floating (pun intended) in my head since I heard about Japanese design studies about foldable aircraft carriers (actually it is just the runway which is expandable and thin).
In the age rockets self-landing on a small sea-born platform, I assume that we have the wave movements under control, at least in principle.
Related

Is there any role today that would justify building a large single dish radio telescope to replace Arecibo?
Could we carve a large radio dish in the Antarctic ice?

References

Osaka University: Department of Naval Architecture and Ocean Engineering
Wikipedia on the Drilling platform "Erik Raude" (in German). This shows that even in rough seas, massive vessel canbe kept extremely stable and static.

I would be extremely concerned about the ability of such a telescope to make adequately precise measurements, given the motion of the water. The leading radio telescopes have their mirrors and receivers very exactly aligned. For example, Arecibo's Gregorian dome could be aligned with any location on the order of millimeters, while the Green Bank Telescope's mirrors are aligned to the tenth of a millimeter. I would certainly be concerned about replicating this performance in the ocean. It's fine for an aircraft carrier's surface to wobble by a couple orders of magnitude more than that; it's less so for a world-class radio telescope. I would expect to see mirror deformation and pointing errors due to the motion of the water.
There's a basket of additional logistical issues - constructing the telescope on water, transporting personnel to and from the structure, combating temperature variations, protecting electrical components from short-circuiting, staving off large waves, etc. All of these drive up construction and operating costs, and astronomers aren't exactly rolling in cash.
I can admittedly see two possible benefits from the plan. First, radio frequency interference (RFI) from artificial sources would presumably be nonexistent throughout much of the open issue (although Connor Garcia makes the excellent point that the increased amount of water vapor could be problematic!). RFI is the bane of the radio astronomer's existence, which is why radio quiet zones are so important. As a counter point, of course, there are available formal and natural radio quiet zones which could be taken advantage of. Second, you could change the telescope's latitude, which would give it a wider range of sources. On the other hand, it's not as if any significant parts of the radio sky can't be seen by radio telescopes, so it's unclear as to the benefits of this.
Overall, I'd say that the precision issues outweigh the potential benefits. If the National Science Foundation were to fund future replacements for Arecibo, there are plenty of better options, some of which are already being planned. If an aquatic radio telescope were ever to be built, I'd guess that it would be entirely privately funded by a billionaire with money to spare who would be willing to take significant risks. I wouldn't complain about something like that, but at the same time . . . I think a conventional radio telescope would be a much better choice.


What exactly is the accretion disk? Is it really possible for material such as dust, gas and other stellar debris that come near the black hole, doesn't fall into it, instead forms a flattened band of spinning material?

A basic answer:

To us on Earth, it looks like everything falls mostly straight down and hitting the Earth.   We almost never see something falling, trapped in Earth's gravity and missing the Earth but in space that happens a LOT.

If we lived in space our observation would be very different.   Most of the stuff we see flies around, in orbits, around the Sun or Earth without colliding into them.  The reason for this is tangential velocity.  It's actually hard for something to hit something else in space, even with gravity, because space is (shockingly) mostly empty space.

Planets, the Sun, black holes, etc, are very small relative to their gravitationally dominant regions or hill spheres.

Why is it hard to hit the sun?  (this applies to black holes too, even more so)

Why is the solar system flat?

Mostly, objects caught in the gravitationally dominant region around a black hole will just orbit the black hole.

When an object passes inside a black hole's Roche limit, it will begin to break apart into smaller parts, and those smaller parts into smaller parts, and close enough, pretty much all matter just becomes a bunch of debris, kind of similar to Saturn's rings but much faster moving and more dense.   Ring systems can form inside the Roche limit around any massive body, large moon, planet, sun or star, though stars which emit lots of light and heat aren't good for the stability of ring systems.

Ring systems, kind of like Saturn's, can form around black holes when there is a collision or when an object passes too close and ring systems tend to flatten out over time due to collisions.  The angular momentum remains constant but the up or down motion of objects cancels out.  The flat accretion disk is actually conservation of momentum, it forms naturally, even if it seems like a lot of wasted space, the matter doesn't know where else to go but into a flattened, quickly rotating disk.

Because of the laws of relativity and gravitational waves, inside about 3 Schwarzschild radii, the ring material much more quickly spirals in towards the black hole, moving faster into a smaller region, getting very hot and extremely magnetic and that's generally what's called the accretion disk.  (Ring systems are generally further out and not hot or spiraling in, though they are loosely similar)

All happens outside the black hole.   Nothing that happens inside the black hole gets out, but what happens outside the black hole in the accretion disk can be surprisingly bright and energetic.  Brighter than the brightest stars.

It might seem like stuff should just fall into a black hole like a vacuum sucking things in, but in gravity in space doesn't work like that.  Anything with zero tangential velocity will fall straight in (kinda-sorta, there's also the dragging space around it problem for rotating black holes), but for simple non rotating black holes, Only an object with close to zero tangential velocity will fall straight, but almost nothing that moves through space has zero tangential velocity relative to it's nearest large gravitational body.  This applies to dust, gas, comets, meteors, planets, stars, anything that passes close to or orbits a black hole.

That's the overall gist of it.  Corrections welcome.