Can Mercury's core "boil over" if it gets to close to the Sun? I was reading about how Mercury's core makes up most of the volume of the planet. I'm guessing this is both because of the small size and the distance to the Sun. If Mercury got too hot could it completely "liquidify"? Would it join the Sun and completely merge with it, gradually crack apart and spread, or something else entirely? Would it cause other problems in terms of life on Earth?

I was reading about how mercurys core makes up most of the volume of the planet. Im guessing this is both because of the small size and the distance to the sun.

There are at least two hypotheses about the causes of Mercury's composition:

Early in the Solar System's history, Mercury may have been struck by a planetesimal of approximately 1/6 that mass and several thousand kilometers across. The impact would have stripped away much of the original crust and mantle, leaving the core behind as a relatively major component.
Alternatively, Mercury may have formed from the solar nebula before the Sun's energy output had stabilized. It would initially have had twice its present mass, but as the protosun contracted, temperatures near Mercury could have been between 2,500 and 3,500 K and possibly even as high as 10,000 K. Much of Mercury's surface rock could have been vaporized at such temperatures, forming an atmosphere of "rock vapor" that could have been carried away by the solar wind.

But if the second hypotheses is right Mercury should have been relocated to wider orbit later. Planetary migration is very interesting topic and I would say it's not solved completely yet.
At current orbit Mercury is not hot enough to evaporate silicates at any significant rate.

If mercury got too hot could it completely "liquidify"?

A lot of "hot jupiter" exoplanets have benn observed in recent years and results are very interesting. For example HD 209458b "Oziris" has a comet-like tail of evaporating atnosphere. Clouds of iron and titanium dioxide were observed too. Iron rain (literally) could be one of exciting effects.
Compared with "hot jupiters" Mercury is much smaller with less gravity. If closer to Sun it undoubtedly would have comet-like tail of evaporating elements, mostly silicates and metals. If intensive enough the tail could be seen by eye with black glass during Mercury transits (also the transits would be much more often at closer orbit).
At closer orbit Mercury should be tidally locked. The Sun-faced surface would be completely melted lava, but dark side would probably be solid because of lack of stable atmosphere. Evaporated rocks would cool down by thermal radiation and dowfall behind the terminator line.

Would it join the sun and completely merge with it, gradually crack
apart and spread, or something else entirely?

The planet should be affected by some force to migrate even close and close to Sun. I can't tell surely will evaporation slow down the planet and move it closer to Sun but I suppose it will not. I suppose Mercury will not fall to Sun but will be evaporating. I can't do math now to estimate how close to Sun it should be to evaporate completely in less than 5 billion years (after that Sun will become a red giant and will envelope Mercury by its atmosphere in "coup de grace" act :) ).
I don't see reasons for Mercury to crack apart if it is not close to Roche limit (again with assumption that evaporation only is not enough to lower its orbit).

Would it cause other problems in terms of life on earth?

I don't think so. Maybe Solar system space would be more dusty because of Mercury's evaporating tail. Even cracking of Mercury would not affect Earth because the debris would stay close to Mercury's orbit (much closer than current Mercury's orbit, don't forget!). Theoretically some of debris pieces could develop more eccentric orbits in millions of years  to reach Earth (because of collisions, uneven evaporation, orbital resonanses) but I bet all them would evaporate before that.
P.S. Most of this answer consists of educated (or not so educated) assuptions. So objections, amendments, corrections are welcome.


Can a stellar mass black hole swallow Earth? In the Astronomy magazine I read that the stellar mass black holes are only the size of Rhode Island. Can they swallow Earth?

There's a fairly common misconception in this question that black holes "suck everything in".   They do - at distances close to their event horizon but as noted, the event horizon's diameter is about the distance across rhode island (different masses have different sizes), but many stellar mass black holes would fit inside the border of rhode island.  Virtually all of them would fit inside pennsylvania for example.   They're tiny.    I'd rate my chances of survival as much higher in a space-ship 1 million miles from a stellar mass black hole than in a space-ship 1 million miles from our Sun.

At normal planetary distances, a planet would just orbit a black hole and likely maintain a stable orbit for billions of years.   Gravitationally, they're no different than a star at those distances.  The orbital mechanics is based on mass and a stellar mass black hole has stellar mass.   What makes it different is it's shrunk inside it's schwarzchild radius, so compared to stars, it's incredibly small and very dense.   The "weird stuff" only happens at very close distances - basically at distances you can't reach around a star because you'd be inside the star.

So . . . what would happen if you put the Earth next to a black hole.

I'm not sure why you'd want to do that, but the Earth would begin to break apart form tidal forces before it got that close.  The gravity would be stronger on the near side than the far side and the Earth would stretch.  This is called spagettification.

If you theoretically made a black hole just appear near the Earth then it would begin to eat the Earth.   Think of a black hole as a drain, a drain some 20-30 miles across, and think of the Earth as a liquid trying to get down that drain.   It's a lot of material squeezing into a comparatively tiny space and in the process the Earth material would get very hot, and likely form into an accretion disk trying to get inside the black hole and the formation of the accretion disk would eject two cosmic jets along the axes.   Black holes are sloppy eaters.

I couldn't tell you the precise amount of Earth that would be absorbed if they were placed right next to each other, but that's also an impossible scenario.   Everything in space is moving and gravity guarantees that if the Earth did meet or closely orbit a stellar mass black hole, that it would be moving enormously fast.

Earth could in theory, but extremely unlikely, either orbit a black hole or collide with one, or pass very close to one, close enough to break apart.  Each scenario would have a different outcome with a different percentage of Earth falling into the black hole.

I don't feel like going through each scenario individually, but the short answer is that the Earth would get quickly ripped apart and some of it, but not all, would fall into the black hole.  Some would be ejected as high speed charged particles.   The matter spiralling in would create a very strong magnetic field and because protons and electrons are charged, much of the matter wouldn't entirely follow the gravitational forces.  Some would get shot out as jets of charged particles.

That's a rough answer anyway.


Reference for broader spectral lines? I'm looking for some reference listing spectral lines outside the visible spectrum.  (In particular, I'm looking for an element or compound that emits strongly in the 700-900nm range.)

Assuming your question is a reference request:

I now about one resource that has a very broad range of spectral data, the

NIST Atomic Spectra Database Lines Form

This has worked great for me.

This page lists a couple of different alternatives. For the near IR case you are interested in, several collections are listed here.


Can somebody explain heisenberg uncertainty principle to 14 year old me? I'm a young student that's very curious about physics and especially some physics theories. I've watched a TED video about the Heisenberg uncertainty principle, but I couldn't fully understand it and I'm looking for a fairly easy explanation if possible.

I've watched a TED video about the Heisenberg uncertainty principle, but I couldn't fully understand it and I'm looking for a fairly easy explanation if possible.

A comment points to Can the Heisenberg Uncertainty Principle be explained intuitively? and notably there are 17 answers there! It is possible that a few will be helpful to any given person.
Quantum mechanics is a set of theories and equations that predicts remarkably well what happens. That doesn't mean that anybody actually understands it. Many people know how to use it, and many many more people kind-of know roughly how to use it. And at least 17 (if not 17,000) people think they know how to explain it.
But don't let them fool you.
Take it from the master explainer. Per Richard Feynman:

Nobody understands Quantum Mechanics!

They behave in a way, that is like nothing you've seen before! Your experience with things you've seen before is inadequate, incomplete...

The behavior of things that are small is so fantastic, is so wonderfully… different. So marvelously different than anything that behaves on a large scale.


Does a black hole necessarily contain a singularity? I'm basing this hypothesis on the theories around how an outside observer watching something fall into a black hole will never actually see the thing pass the event horizon. And secondly that from the perspective of the thing falling in, the surrounding universe would accelerate to see the end of the universe.

On that second point, the "end of the universe" from my understanding would be up until the black hole evaporates.

So the question. Is it possible for an event horizon to form; not necessarily from a singularity, but merely from there being enough compressed mass existing within the boundaries that an event horizon forms?

The implication from this if it has any merit would be in my opinion, that there are no singularities currently in existence and that there indeed might never be one, because currently I don't understand how something "inside" a black hole is capable of moving at all (I assume that to do so mass would need to move faster than light), suggesting that the entirety of a stellar remnant's mass is stuck in place both inside and on the surface of the event horizon.

EDIT: Re-reading that last paragraph I've confused myself a little, mass wouldn't necessarily be "stuck", but if it did move at all, it would be moving light speed, which is a paradox in itself?

Yes, an event horizon will form once escape velocity exceeds the speed of light. It does not necessarily require a singularity inside it.

However, there is no known physics that would allow any non-point-like object to exist inside an event horizon. It is hard to see how any traditional forces would be able to support an object in such case, and for quantum mechanics we simply do not have the theory that'd be valid in such high gravitational field.


How does neutron star collapse into black hole? We know the spectacular explosions of supernovae, that when heavy enough, form black holes. The explosive emission of both electromagnetic radiation and massive amounts of matter is clearly observable and studied quite thoroughly. If the star was massive enough, the remnant will be a black hole. If it wasn't massive enough, it will be a neutron star.

Now there's another mode of creation of black holes: the neutron star captures enough matter, or two neutron stars collide, and their combined mass creates enough gravity force to cause another collapse - into a black hole.

What effects are associated with this? Is there an explosive release of some kind of radiation or particles? Is it observable? What physical processes occur in the neutrons as they are subjected to the critical increase of pressure? What is the mass of the new black hole, comparing to its neutron star of origin?

A neutron star must have a minimum mass of at least 1.4x solar masses (that is, 1.4x mass of our Sun) in order to become a neutron star in the first place.
See Chandrasekhar limit on wikipedia for details.

A neutron star is formed during a supernova, an explosion of a star that is at least 8 solar masses.

The maximum mass of a neutron star is 3 solar masses. If it gets more massive than that, then it will collapse into a quark star, and then into a black hole.

We know that 1 electron + 1 proton = 1 neutron;

1 neutron = 3 quarks = up quark + down quark + down quark;

1 proton = 3 quarks = up quark + up quark + down quark;

A supernova results in either a neutron star (between 1.4 and 3 solar masses), a quark star(about 3 solar masses), or a black hole(greater than 3 solar masses), which is the remaining collapsed core of the star.

During a supernova, most of the stellar mass is blown off into space, forming elements heavier than iron which cannot be generated through stellar nucleosynthesis, because beyond iron, the star requires more energy to fuse the atoms than it gets back.

During the supernova collapse, the atoms in the core break up into electrons, protons and neutrons.

In the case that the supernova results in a neutron star core, the electrons and protons in the core are merged to become neutrons, so the newly born 20-km-diameter neutron star containing between 1.4 and 3 solar masses is like a giant atomic nucleus containing only neutrons.

If the neutron star's mass is then increased, neutrons become degenerate, breaking up into their constituent quarks, thus the star becomes a quark star; a further increase in mass results in a black hole.

The upper/lower mass limit for a quark star is not known (or at least I couldn't find it), in any case, it is a narrow band around 3 solar masses, which is the minimum stable mass of a black hole.

When you talk about a black hole with a stable mass (at least 3 solar masses), it is good to consider that they come in 4 flavors: rotating-charged, rotating-uncharged, non-rotating-charged, non-rotating-uncharged.

What we would see visually during the transformation would be a hard radiation flash.
This is because during the collapse, the particles on/near the surface have time to emit hard radiation as they break up before going into the event horizon; so this could be one of the causes of gamma ray bursts (GRBs).

We know that atoms break up into protons, neutrons, electrons under pressure.

Under more pressure, protons and electrons combine into neutrons.

Under even more pressure, neutrons break down into quarks.

Under still more pressure, perhaps quarks break down into still smaller particles.

Ultimately the smallest particle is a string: open or closed loop, and has a Planck length, which is many orders of magnitude smaller than a quark. if a string is magnified so it is 1 millimeter in length, then a proton would have a diameter that would fit snugly between the Sun and Epsilon Eridani, 10.5 light years away; that's how big a proton is compared to a string, so you can imagine there are perhaps quite a few intermediate things between quarks and strings.

Currently it looks like several more decades will be needed to figure out all the math in string theory, and if there is anything smaller than strings then a new theory will be required, but so far string theory looks good; see the book Elegant Universe by Brian Greene.

A string is pure energy and Einstein said mass is just a form of energy, so the collapse into a black hole really breaks down the structure of energy that gives the appearance of mass/matter/baryonic particles, and leaves the mass in its most simple form, open or closed strings, that is, pure energy bound by gravity.

We know that black holes (which are not really holes or singularities, as they do have mass, radius, rotation, charge and hence density, which varies with radius) can evaporate, giving up their entire mass in the form of radiation, thus proving they are actually energy. Evaporation of a black hole occurs if its mass is below the minimum mass of a stable black hole, which is 3 solar masses; the Schwarzschild radius equation even tells you what the radius of a black hole is given its mass, and vice versa.

So you could transform anything you want, such as your pencil, into a black hole if you wanted to, and could compress it into the required size for it to become a black hole; it is just that it would immediately transform itself (evaporate) completely into a flash of hard radiation, because a pencil is less than the stable black hole mass (3 solar masses).

This is why the CERN experiment could never have created a black hole to swallow the Earth - a subatomic black hole, even one with the mass of the entire Earth, or the Sun, would evaporate before swallowing anything; there is not enough mass in our solar system to make a stable (3 solar mass) black hole.

A simple way for a neutron star to become more massive in order to be able to turn into a black hole is to be part of a binary system, where it is close enough to another star that the neutron star and its binary pair orbit each other, and the neutron star siphons off gas from the other star, thus gaining mass.

Here is a nice drawing showing exactly that.

Matter falling into a black hole is accelerated toward light speed. As it is accelerated, the matter breaks down into subatomic particles and hard radiation, that is, X-rays and gamma rays. A black hole itself is not visible, but the light from infalling matter that is accelerated and broken up into particles is visible. Black holes can also cause a gravitational lens effect on the light of background stars/galaxies.


How did early estimates of a "potato radius" set 1 eV ~ GMμ/R and get 200 to 300 km? @DavidHammen's answer explains that Goblin has roughly a potato radius and I can't believe I just wrote Goblin and potato in a serious question. That happened.
The linked paper; The Potato Radius: a Lower Minimum Size for Dwarf Planets (Lineweaver & Norman 2010) says:

Derivation of the Potato Radius
One could begin, as the author did initially, by setting the gravitational acceleration at the surface of a body equal to the acceleration needed to undo the electronic bonds. But this is not what we want since it is equating surface gravity with the force needed to crush rocks. The Earth is spherical but its gravitational pull is not so strong that it crushes rock at the surface. One has to go down ~10 or ~20 kilometers before the overburden pressure can do that.
To our knowledge, the published derivation of something most analogous to the potato radius is in Chap. 1.2.1 of Stevenson (2009). Focusing on energy (not force or pressure) the electronic bond energy of ~ 1 eV is set equal to the gravitational energy ~ GMµ/R where µ is the mass of the particle in question and R is the radius of the object. The result is a radius of a few thousand kilometers for a rocky body and ~ 1000 km for an icy body. This was meant to be only an order of magnitude calculation. We try to improve on it below.

We have there
$$1 \text{eV} \approx \frac{GM\mu}{R} $$
1 eV is about 1.6E-19 Joules and $G$ is the gravitational constant, but how to I use this to arrive at a potato radius $R$ of about 200 to 300 km? I can read the words in the quote, but I don't understand how to proceed.

How do I use this to arrive at a potato radius R of about 200 to 300 km?

You can't, for multiple reasons. One reason is that the authors themselves disclaim this as a valid approach:

  The result is a radius of a few thousand kilometers for a rocky body and ~1000 km for an icy body. This was meant to be only an order of magnitude calculation. We try to improve on it below.

Another is that the authors don't show their math. On trying to replicate it, I get rather different results. The mass of a more or less spherical object is $M=\frac43\pi\rho R^3$, where $\rho$ is the mean density of the object. Plugging this into their expression yields $1\,\text{eV}\approx \frac43\pi G\rho\mu R^2$, or
$$R \approx \sqrt{\frac{1\,\text{eV}}{\frac43\pi G\rho\mu}}$$

This suggests the potato radius is inversely proportional to the square root of density and inversely proportional to the square root of particle mass. This obviously is not the case; the observed potato radius of rocky objects is about 50% more than that of ice objects, even though rocky objects have the greater density and are made of more massive molecules.

I took this as a throwaway paragraph. The key results are further on in the article.


Can we change Earth's orbit with modern technology? It is well known that, according to Newton's 3rd law, spacecrafts entering or leaving Earth provoke tiny (insignificant) changes to Earth's orbit (see related question here).
My question is about significant changes. For instance, say we would like to shift the Earth's orbit closer to that of Mars (forget about the ecological consequences of this). Do we currently have the technology capable of doing this? For instance, something like a huge nuclear explosion? This question only deals with "natural", non-human ways to change a planet's orbit. I am interested about technology-based changes.

No, we cannot.
It's relatively simple math to show that we are a very long shot from changing Earth's orbit by anything significant at all: consider the kinetic energy of Earth in its current orbit, and do the same math for that in the desired orbit:
$ E_{kin} = \frac{m_E}{2}v^2 \approx \frac{m_E}{2}\frac{GM_S}{a}$
where $m_E = 6\cdot 10^{24}$kg is Earth's mass, $M_S=2\cdot 10^{30}$kg is the solar mass and $a=150\cdot 10^9$m is the distance from Earth to sun. I used the approximation in calculating orbital velocity $v = \sqrt{\frac{GM_S}{a}}$ that the Earth mass, is much smaller than the Sun's mass. So current orbital velocity is (using numbers as above) $29822$m/s.
Now, changing oribital distance outward by 1%, thus by $1.5\cdot 10^9$m, to $151.5\cdot 10^9$m yields a speed of $29674$m/s - and in energy that subsequently gives a difference of $2.64\cdot 10^{31}$J to move Earth's orbit about 1% to further from the Sun.
Annual world primary energy production currently is around $10^{20}$J - so we "just" need to produce 100.000.000.000 times the yearly energy production of the entire world to change Earth's orbit about 1%. And that implies that we have a technique to transfer energy 100% into Earth orbital velocity - something which cannot exist either due to simple thermodynamic argument.
Edit to add: Another way to argue is the rocket way: typical speed of conventional rocket exhausts are a few km/s, which is less than needed to leave Earth's orbit - that's why you need multi-stage rockets to get anything into orbit. But if you want to move the whole Earth with a rocket motor, you need a propellant which is fast enough that it directly leaves Earth's gravitational field - or you will not be able to create any effective change of momentum and only move around mass in the Earth system itself. So you need to make some substantial technological advances here, too - or some entirely different technique.


How to calculate local position, knowing the RA and Dec of the center of an image, RPY of the camera, and precise time? Given a picture of stars - assume it is possible to identify the stars and their Right Ascension (RA) and Declination (Dec).  By choosing one near the center of the image (or interpolating RA/Dec to the center) the effective "pointing angle" of the camera would be known.

With the RA/Dec of the camera, plus the Roll (R), Pitch (P), and Yaw (Y) of the camera, and the precise time (TAI, UTC, and/or UT1) the picture was taken... how would a location calculation of where the camera was on Earth when it was taken be possible (assume it is possible).

It seems like it would be a simple coordinate transformation, however understanding how time affects the calculation isn't clear.  I can't find a simple resource of algorithms to find one close enough.

ETA: Roll, Pitch, and Yaw could be found in a few different ways... in my current case I took a good first initial "guess" and then plotted an overlay of stars based on perturbing the initial first guess until the stars overlaid well over the image (of course, this required I knew where the picture was taken!)  However, it is also possible to know R,P,Y based on a local gravity vector and inertial mechanics (a simply INU)... which is where I would eventually like to get to.

The RA and Dec is currently found using http://astrometry.net/ by uploading the picture they returned astrometric meda-data.

Here is an example set of numbers:
R=10.7deg, P=63.1deg, Y=-174.8deg

Time=2016,Oct,1,0hr,31m,28s

Leap Secs=36 GPS-TAI

UT1-UTC = -0.28

Center of Image (~pointing vector of camera)

RA = 90.7 (6h,2m,53.422s),

Dec=3.56(3deg, 33' 24.95)

Picture was taken somewhere near 30°22'02.0"N 75°00'35.7"E
Bathinda, Punjab, India.
Answering your question about time - how it enters into the sight reduction: RA, all by itself, isn't useful. It's the angle between the plane of the meridian of a celestial location and the plane of the meridian of the first point of Aries. That angle is more or less fixed, but the first point of Aries, and its meridian, are continually changing, going through 360 degrees in a sidereal day.
Assumptions: pitch is the altitude of the celestial location (its angular distance above the horizon), and it's not corrected for refraction. Yaw is the azimuth of the location, CW degrees from north.
General method: knowing the time, find the Greenwich hour angle (GHA) of the first point of Aries at that time, by interpolating between hourly values in an almanac. The GHA of the celestial location is the GHA of Aries minus RA. The geographical position (GP) of a celestial location, is the point on the earth's surface where the celestial location is in the zenith. The GP I found after some number crunching was 3.56 deg N, 72.655 deg E. It's in the middle of the Arabian Sea. That GP, the observer's unknown location, and the north pole form a spherical triangle, which must be solved to get the observer's location fix. Two sides and an angle are known: one of the sides is 90 deg minus the latitude of the GP, the other is 90 deg minus the observed altitude of the celestial location. The angle is the 360 deg minus the azimuth of the location. It's one of those awkward spherical triangles where the angle is not the included angle between the sides. It's solved by first using the law of sines to find the angle at the pole then using Napier's Analogies to find the third side.
In this particular case, the celestial location GP is almost due south of the observer, so there's a way to make a simple estimate. The observer is roughly 26.9 deg north of the GP, a little less than 30.46 deg N, and a little east of the GP 72.65 E.
I have a few more things to say if anyone's interested. I'd appreciate a check on my numbers. It's very easy to make a mistake, which is why sight reductions in the pre-GPS era were done by two persons independently who'd compare their results. An unintentional grounding can ruin your whole day.


How dependent is Earth is with the other Planets' gravity? How would the Earth be affected if the Moon was destroyed as far as our orbit around the Sun? What if Mars or Jupiter was destroyed?

The orbital speed of the Moon around the Earth is 1022 m/s. The Earth is 81 times more massive, than the Moon. Thus, a sudden disappearance of the Moon would cause an $\frac{1022}{81}=12 \frac{m}{s}$ change of the Earth's orbit, which would be negligible.

The disappearance of the Mars or Jupiter would have yet lesser effect.


What are the biggest problems about the numerical, finite-element GR models? As I know, for example the modelling of the collapse of a neutron star (to a black hole) wasn't done correctly until now.

Why? Yes, I know, the Einstein Field Equations aren't really easy to solve. But, for example, the situation is very similar in the case of the QCD, but there are already fruitful results existing ( http://physicsworld.com/cws/article/news/2008/nov/21/proton-and-neutron-masses-calculated-from-first-principles ).

Is there any theoretical obstacle in the way? Or we simply didn't have enough fast computers/physicist/programmer manyear until now?

If you can provide examples of numerical methods in GR you've seen/heard of that would help focus the question.

From the article you linked to: "The technique keeps track of a vast number of quarks and gluons by describing the space and time inside a proton with a set of points that make up a 4D lattice". This almost gets to the main issue with Numerical Relativity. There is no natural computational grid on which to simulate space-time. The whole game with GR is that gravity is space-time so first you have to simulate the space-time and then you have to simulate the objects (neutron stars, black holes, gravitational waves) on top.

As the links below go into, its very difficult to create a consistent computational grid since the physical space-time your trying to simulate for a black hole has "funny" things in it like singularities, or an event horizon pas which we can't really know what's going on.

I think this article: http://astronomy.com/magazine/2016/02/putting-einstein-to-the-test?page=1

does a good job of summing up the field, and its quite accessible.

For something more rigorous please see: http://arxiv.org/pdf/1010.5260v2.pdf
That paper gets into some of the math behind the article linked to above.


Do bigger telescopes equal better results? There are many large reflecting/optical telescopes around the world, just to name a few: the Very Large Telescope, Large Zenith Telescope, and European Extremely Large Telescope.

My question is does actually building larger telescopes improve visibility of the universe, quality of an image, and various other attributes of a telescope?

Well it does seem kind of obvious that a larger telescope may have more magnification and better imaging, however if that is true than why don't we just build a super large one (think really really big, unlike the size of the current large ones) so we can see the universe in HD - so to speak?

Since we haven't done this yet (I think) is it because of financial issues or is it because simply building a bigger telescope won't improve much to visibility and image quality?

Magnification is not the main purpose of a telescope. A very small telescope will have a lower resolution than a larger one, but for larger ones, the atmosphere limits the amount of detail visible.

Large telescopes capture more light than small telescopes. More light means that dimmer objects are visible, and there is more light to be analysed. A very dim object, such as a planet orbiting another star, needs a very large mirror to collect enough light to analyse spectroscopically. And this is the reason why larger telescopes are made. Of course larger telescopes cost a lot, and cost is the reason that telescopes like the "Overwhelmingly large telescope" was not built.

Much of the development in telescopes in the last 25 years has been not in ever larger mirrors, but in adaptive optics, or using two telescopes together as an interferometric telescope.


Do astronomers have an established, systematic way for saying what does or doesn't orbit what? (e.g. "Mars orbits Earth") A recent comment

  An object far enough away can certainly orbit the Moon and the Earth (and the Sun) -- Mars, for instance does this. An object in the Earth-Moon L2 is also orbiting both the Earth and the Moon.

A recent comment about that comment:

  according to your way of thinking all solar system bodies orbit Mercury, except the Sun, which is just silly. Or does the Sun in fact orbit Mercury as well? I think your use of the word orbit is not workable, nor shared by almost anyone else.

So I'd like to ask if astronomers have an established, systematic way for saying what does or doesn't orbit what within the solar system, and if so, a link to point to in the future would be greatly appreciated!

It is possible that there is nothing "official", there is just the technical use of language. For example Phil Plait notes that He incorrectly used the word "orbit" for the motion of the Hayabusa probe, which does not orbit the asteroid Ryugu. But hovers over the surface.

So what does Phil mean when he talks about orbiting?  I summarize the meaning thus: Object A is in orbit around object B if A moves around B (ie the True Anomaly of A increases from 0 to 360 degrees in fairly regular fashion) principally as a result of the gravitational field of B.

So although Mars does move around the Earth, its motion is not principally due to the gravitation field of the Earth, so it is not in orbit.  However the moon's motion is principally due to the gravitational field of the Earth. It is in orbit around the Earth.

The moon is also in orbit around the sun: It moves around the sun and the reason for this motion is principally the graviational field of the sun.

A body at the Earth Sun L2 point is not in orbit around the Earth (it is fixed relative to the Earth) It is in orbit around the sun. A body in the Earth Moon L2 point (which is probably very unstable) is orbiting the Earth and orbiting the sun. Its motion relative to the Earth is primarily due to the Earth (and secondarily due to the moon). Its motion relative to the sun is due to the gravity of the sun.

Cruithne is in a 1:1 resonance with the Earth, but its postion is not primarily due to the gravity of the Earth. It is in orbit around the sun, not the Earth.

Hayabusa is not in orbit around an asteroid, it is maintaining position by thrusters. Similarly, A Jumbo jet is flying not orbiting, it's motion is not mainly due to gravity.

This idea can be sharpened by the notion of a Hill sphere.
Inside the Hill sphere, a body will orbit. Outside it, the body will not orbit.


Are ascending node of sun, the point of intersection of prime meridian and equator and center of earth all collinear? I did a bit of googling but couldn't find the exact answer, so are the following 3 points collinear or not ?

1_center of earth

2_ascending(or descending) node of sun

3_point of intersection of prime meridian and equator(point of zero longitude on equator).

Not exactly.

During either of the equinoxes there is a moment when the line between center of earth and the sun aligns with the equator. This doesn't coincide with prime meridian in any way though; it may happen at any meridian whatsoever that happens to coincide with the line.

Of course if it happens the sun is in zenith above prime meridian at that moment, your conditions will be satisfied. But depending on precision you allow, that may be a very short time window. It's all about how precisely you want that to be.

Near Equinox Earth tilts by about 0.25 degree of latitude per day, meaning that is the maximum angle by which the line will be off from equator while crossing the prime meridian on the same day as when Equinox happens. 0.25 degree of latitude is about 27 kilometers, so this is the maximum error. Of course on some years, that will be much less, it's just that Earth tilt is correlated with length of year, zenith line crossing of prime meridian is correlated with time of day, and time of day is not really correlated with length of year.


Can we detect pulsars that aren't "pointed" towards us? A pulsar is a highly magnetized rotating neutron star that emits beams of radiation through its magnetic poles, in the direction of the Earth. This makes it seem to pulse, as seen from Earth. There likely are similar objects that pulse as seen from other directions, but not from Earth. Can we tell the difference between one of these objects and an ordinary neutron star?

Is there any way of detecting pulsars that are not pulsing in our direction?

They don't need to be pointed exactly at us, since the angular beam radius is a few degrees (Pulsar Astronomy, P. 212) but there is a portion that we can't detect, described by a beaming factor (P. 211):

The larger beamwidth at higher energies can be seen in these graphs:

R. N. Manchester

We mostly can't see "ordinary neutron stars". "neutron stars can only be easily detected in certain instances, such as if they are a pulsar or part of a binary system" - Wikipedia

If the pulsar is radiating gamma rays more isotropically (but still varying with phase as in the Manchester charts), then these could be detected.


Jupiter orbiting time in Earth's orbit? Imagine if Jupiter orbited the Sun at the Earth's orbital distance of 1AU.

Would a planet of this size orbiting the Sun take 1 Earth year (365 days) to complete an orbit, or would the size of Jupiter affect its rotation and orbit?

If Jupiter was at 1 AU from the Sun, its orbit about the Sun would be about 4 hours and 10 minutes shorter than a sidereal year. That's an effect, but it's not very much of one, due to Jupiter's mass being about 1/1000th of the Sun's mass. For a more significant effect, if our Moon was much smaller than it is, it would orbit the Earth about 4 hours and 1 minute longer than a sidereal month. The Moon's mass is about 0.0123 Earth masses.

The period of the Keplerian orbit of an object of negligible mass about some massive object is $$P = 2\pi\sqrt{\frac{a^3}{GM}}$$ In the above, $a$ is the semimajor axis length, $G$ is the Newtonian gravitational constant, and $M$ is the mass of the central object. When the orbiting object has non-negligible mass $m$, the above expression needs to be modified to $$P = 2\pi\sqrt{\frac{a^3}{G(M+m)}}$$


How do I calculate B-V to make a Hertzsprung Russell diagram? I am reading my textbooks and a few "Astro101" websites, but I cannot find a final answer to this:

In the $B-V$ colour index that is plotted in an H-R diagram, are $B$ and $V$ the (observed) apparent magnitudes, or should they be corrected for extinction first, i.e. with known $A_V$ and $A_B$ values?

Unless you know all the stars in the diagram suffer the same extinction, then you should de-redden them individually.

If you do know they have the same extinction (e.g. they are all nearby stars and it is zero; or they are all stars well above the Galactic plane in a similar direction; or stars in a cluster at the same distance) then you can plot a diagram using the observed magnitudes. In the latter scenario of a cluster, this diagram can be used to estimate the reddening of the cluster.

If you want to compare with models, to estimate temperatures or masses, then you must de-redden (or redden the model!)


How would I search within an angular distance of an object corresponding to a distance in parsecs? I'm interested in the learning about the local densities around galaxies. I've found myself a bit confused on how to relate angular distances (arcseconds) with physical distances (parsecs) when thinking about distances between two distant galaxies. For example, let's say the average distance between galaxies within clusters is a few megaparsecs. If I had ra, dec coordinates of a particular galaxy and could hypothetically query an all-sky catalog, let's say I'd want to search within a few megaparsecs to get a feel for its density field. But generally when querying catalogs you specify a search radius in some arcseconds. How would these relate here?

Ken G's answer is essentially correct, but there is one important thing to keep in mind: The distance $r$ between the galaxies that is interesting for you is the distance they had at the time they emitted the light you see (you don't care about what has happened to them since that time and how far apart they are today; you care about the physical properties then). But the angle you observe them to span is not angle they spanned when they emitted the light (since they were closer to you in the past, so the light they emitted in your direction at that time would "miss" you by the time it reached that place).

This means that, in your calculation, the distance from you to the galaxies should not be the "physical" distance $d$ (i.e. the actual distance right now to the galaxies, which also corresponds to the comoving distance), but the so-called angular diameter distance, $d_A$. These are related (in a flat universe) by
$$
d_A = \frac{d}{1+z},
$$
where $z$ is the redshift of the galaxies (I assume they have almost the same redshift; if they are too different, then the distance from each other is not simply given by the projected distance). Note that in a non-flat universe, the formula is slightly more complicated; see the link above.

The comoving distance, in turn, is obtained by integration:
$$
d(z) = \frac{c}{H_0} \int_0^z \frac{dz'}{\left( \Omega_\mathrm{M}(1+z)^3 + \Omega_\Lambda \right)^{1/2}},
$$
where $c$ is the speed of light, and $\Omega_\mathrm{M}$, $\Omega_\Lambda$, and $H_0$ are the present matter and dark energy density parameters and Hubble constant, respectively. Here, I've again assumed a flat universe, and neglected radiation since when that was important, there were no galaxies anyway).

Obviously, this effect is not big for small redshifts, but $d_A$ has the interesting property that it increases with physical distance only out to a certain point (roughly $z=1.6$), after which it decreases, because the galaxies were closer to us in the past.

So the physical distance $r$ between two galaxies spanning an angle $\theta$ is
$$
r = \frac{\theta}{d_A}
$$
or
$$
\boxed{
\frac{r}{\mathrm{kpc}} = \frac{\theta / \mathrm{arcsec}}{d / \mathrm{Mpc}} (1+z)
 \, \frac{10^{-3} \times 180\times3600}{\pi},
}
$$
where the last factor accounts for converting Mpc to kpc and radians to arcseconds.

In Python, arbitrarily for $z=2.2$ and $\theta=12.3''$,  it's as easy as

from astropy.cosmology import Planck15
from astropy import units as u
z     = 2.2                               # redshift of galaxies
theta = 12.3 * u.arcsec                   # angle
r_ang = Planck15.kpc_proper_per_arcmin(z) # phys. dist. per angle
r     = r_ang * theta                     # physical distance
print('Distance between galaxies: ', r.to(u.kpc))

which will print Distance between galaxies:  104.22540165050975 kpc.


how far away can close stars diverge with time? Do high velocity stars decelerate if they transit at 300ly per million years? What kind of forces can decelerate them and in what kind of time frame?

What about ordinary stars which travel at 70 light years per million, i.e alpha centauri? How often do they change direction and do they end 1000ds of light years away after a billion years?

Only one force counts on galactic scales: gravity.

Stars like Alpha Centauri are orbiting in the galaxy, as is the sun. Both stars are moving in a similar direction and at roughly the same speed, relative to the galactic centre. The distance to such stars will vary slowly, but both stars will remain within the galaxy, and the stars will stay fairly near each other for some time yet. The direction of stars doesn't change suddenly, but the gravity of the galaxy will pull the star's path into a roughly circular orbit within the galactic disc.

High-velocity stars are still affected by the gravitational pull of the galaxy. The galactic escape velocity is over 500km/s (1800 ly per million years) Stars can have sufficient speed to escape from the galaxy's gravitational pull. Such stars can escape from the galaxy into intergalactic space. They will be decelerated by the gravity of the galaxy, but if their speed is enough, then gravity won't be enough to pull them back. Stars that are moving more slowly than that will pass into the galactic halo, before being pulled back by the gravity of the galaxy. They will then usually orbit in the galactic halo, with an orbital period measured in hundreds of millions of years.


Items at the Event Horizon vs. colliding Black Holes Items reaching a Black Hole's Event Horizon seem to be frozen in time from the point of view of an outside observer. On the other hand, Black Holes collide and merge, in finite steps, producing gravitational waves observable to outside observers. How is it possible these two facts not to contradict each other?

I don't know if this question is a good fit for the Stackexchange format, because it's a complex topic that would require a lot of discussion. But I'll try to provide an answer.

I think you're getting the wrong ideas from the image of the event horizon being frozen in time for the remote observer. While this is true, it also means the remote observer never actually gets any information from the event horizon proper.

If you're a remote observer, and you see any change that happens related to a black hole, by definition that change is not happening at the event horizon or inside it. It happens outside that region. The event horizon might as well not even exist from this p.o.v.

The merger of the two black holes is what the remote observer sees. But a hypothetical local observer, inside the event horizon, would have a different perception.

For the local observer, the whole history of anything outside the black hole takes place in an instant. The whole evolution of the universe, including the black hole colliding with another black hole, and the resulting object evaporating via Hawking radiation - all this stuff happens in an instant.

You really need to think of it this way: the remote and the local observer experience two completely separate realities, where the flow of time in one is completely disconnected from the flow of time in the other.


Bimodal Galaxy Distribution Why do red galaxies tend to be more massive than blue galaxies? Do we have an explanation for this observation?

My understanding is that the matter is not settled, definitively. There is a chain of logic that goes something like this:

Why is any galaxy red? Because it isn't forming any new, high mass, short lived, blue stars.
Why would a galaxy not form stars? Because the gas in the galaxy is either not dense enough or not cold enough to gravitationally collapse into new stars.

There are two main explanations put forward for the inability of the gas in really large galaxies to collapse and form stars. First, the most massive galaxies tend to be found in clusters of galaxies, and the gas in the galaxy clusters is too hot to collapse and form stars. Second, when large galaxies of comparable mass merge simulations suggest that the merger process heats up the gas in the galaxies, cutting off star formation.


What will a lunar eclipse look like from moon? What will a lunar eclipse look like from moon? Will earth become a completely dark circle?

It would look like this (actual picture of the Earth, seen from the Moon, during a lunar eclipse):

Link to full page: http://apod.nasa.gov/apod/ap140407.html

The Earth would appear surrounded by a bright ring, even though the Sun is completely hidden behind it. The ring is sunlight refracted through the atmosphere. It's basically all the sunrises and sunsets, seen at once.

The Earth would never appear fully dark. The bright ring around it will always be there. This is due to the atmosphere.


Can I take logarithms of magnitudes (which are already log) I'm investigating the relationship between the magnitude of quasars in various wavelengths, and their redshift $z$. I've found that if I take the $log$-$mod$-transform ($L_{mag}$) of the magnitudes, where

$L(y) = sgn(y) * log(|y| + 1)$

which I got from here and here, I can get a slightly tighter fit for my data, which is the aim of the study. Here's a plot for the W1 magnitude and $L(W1)$ as above, both against $log(z)$:

showing the regression coefficient for each.

My question, before I take this to my supervisor, is: is there any reason why I shouldn't be taking the $log$ of an already $log$ quantity?

You can, the question is whether you should.

Is there a theoretical reason why you think that magnitude should vary exponentially with redshift? If not then you may be falling into a fallacy.

You see with a polynomial with sufficiently many terms, I can get a curve that fits your data perfectly, r=1, Does this my polynomial is a better model? Probably not.  Just throwing functions at your data and choosing the one that gives the biggest r isn't mathematical modelling of data.

I would question why you are using magnitude at all? You seem to be finding that "close objects appear brighter" which isn't a new discovery.


Why you use log to measure metallicity in galaxies? For example, I saw the next expression for matallicity in a paper: $\log(O/H) + 12$.

I understand O/H is the O(Oxygen) to H(Hydrogen) ratio but why is there the number 12? And why the logarithm?

The distribution of metallicities appear to be more evenly spread out in logspace than in linear space. The reason for this can be ascribed to there being no preferred scale for the abundance of a given element; rather they span several orders of magnitude. The same can be said for instance about the distribution of dust grain sizes, the distribution of the mass of dark matter halos, and the distribution of the area of lakes on Earth.

So, if you measure the number of oxygen atoms (or ions) in ten stars, and divide by the number of hydrogen atoms, you may get values like
$$\mathrm{O}/\mathrm{H} = \{0.03,3.5,25,0.003,0.9,0.4,0.09,0.01,8,0.02\}\times10^{-4}.$$
Plotting this in a linear and a logarithmic plot, you see that on a log scale, the values are easier to tell apart:

(Another reason, as zibadawa timmy writes in his/her answer, is that taking the log you don't need to write all those $10^x$ factors).

Now why add 12? This factor corresponds to measuring the number of a given atom per $10^{12}$ hydrogen atoms. I've been asking colleagues from students to professor level, and the best answer we can come up with, is that the least abundant elements in the Sun (uranium, rhenium, thorium, …) have abundances of the order of one per $10^{12}$ hydrogen atoms, so adding 12 to $\log(X/\mathrm{H})$ makes all values positive. This angst of negative values, I don't understand, though. Usually, we're okay with minuses. And measuring metallicities in stars or gas of sub-solar metallicity yield negative values anyway. Moreover, the scale was in use even when we had only lower thresholds for uranium and bismuth of $\log(\mathrm{U}/\mathrm{H})+12<0.8$ and $\log(\mathrm{Bi}/\mathrm{H})+12<0.6$, respectively (Grevesse 1969) and didn't know if they were actually smaller.


Do all stars have the potential to have life supporting planets? Regardless of how big or small, how hot or cold, do all stars have a goldilocks zone, and if so, could any given planet in this zone (with the right configuration) support life?

Great question. The goldilocks zone is usually defined in terms of a region where the equilibrium temperature of the planet lies between some temperature limits (these temperature limits are somewhat debatable, but irrelevant for the purposes of this question - the boundary becomes fuzzy). This region can be calculated by working out how much flux is received from the star at a given radius.

In that sense - all stars have a goldilocks zone - you can always work out some distance from a star where the flux is roughly equivalent to the flux we receive from the Sun for instance. It is much closer to a dim star and much further away from a luminous star.

But, there is a difference between a habitable zone and a continuously habitable zone. If you seriously want life to evolve on a planet, then it will take time. How much time? No-one is sure but it seems to have taken a few hundred million years in our solar system. There are also good reasons to suppose that young planets are not going to be habitable - either they are still extremely hot after formation or they are being bombard by debris.

Thus you can probably exclude any star with a main sequence lifetime of less than about 100 million years (or longer if you feel conservative about how long it takes life to get going). This rules out stars of more than about 5 solar masses.

You might also rule out stars that evolve quickly. If a star changes its luminosity quickly on a short timescale then the habitable zone moves drastically too. This will happen near the ends of the lives of all main sequence stars and for all subgiant and giant stars. So for these, yes there is an instantaneous habitable zone, but no region is in a habitable zone for hundreds of millions of years or more.

You might want to consider not only the luminosity of the star, but its spectrum as well. For instance, both hot white dwarfs and low-mass M-dwarfs are reasonably stable and long-lived; they are faint, so have habitable zones close to the star (much closer than 1 au). However, for different reasons, both these types of object emit copious UV radiation. In hot white dwarfs it would be because they have hot photospheres. In M-dwarfs they can be highly magnetically active into old age, having a hot chromosphere and corona that would strongly irradiate any nearby planet. You might consider that their "habitable zones" were in fact uninhabitable. Of course an atmosphere and a strong magnetic field might counteract this.

A lot of this depends on your terms of reference and definition. The wikipedia page on habitable zones has quite a nice discussion, which emphasizes the debatable nature of this topic.

A final thought, which is not in the wiki page. Multiplicity could mess things up. Planetary systems will not be stable in the habitable zone of either star in a binary system if their separation is comparable with the orbital radius of the habitable zone. If you had two stars like the Sun, their separations would have to be less than a few tenths of an au or greater than of order 10 au in order to allow something to orbit at about 1 au around one of them. Even then, there could be all sorts of dynamical instabilities which prevent long-lived planets, especially if there were other planets in the system too.


Would a direct hit by a small comet be catastrophic? Would a direct hit by a short-period comet with a 3 mile diameter nucleus cause enough damage on Earth to threaten or destroy human life and the ability to produce crops?

Here, have some fun with this...

http://impact.ese.ic.ac.uk/ImpactEffects/

It doesn't go into long term effects, but the short term is pretty devastating for a significant portion of the planet.  'Nuclear winter' effects seem inevitable for those not atomised, scorched, crushed, or covered in ejecta.

Not good, nope.


How close would merging black holes have to be to feel gravitational waves? Recently LIGO discovered gravitational waves caused by two black holes that were orbiting each other, and then collapsed into one black hole. A few months later, we find out that this actually happens quite often. However, these collisions are far enough away from Earth that no one can feel the gravitational waves, and it actually takes very precise measurement to even determine that they happen. So my question is, how close would a black hole collision have to be for us to feel the gravitational waves pass through Earth?

Secondly, is this even possible? As in, would the distance have to be so short that we'd get swallowed up before the black holes collided? And, if we were to feel such gravitational waves, what would they feel like?

EDIT: Let's say that the black holes have the same mass as Earth, giving their event horizons a radius of about 9 mm (the size of a dime). So two Earth-sized black holes that are stuck in an orbit finally collapse is the scenario here.

It's a bit hard to answer this question as there are some ambiguities and certain parameters of the setup aren't defined, but I can try to give you a general idea.

  If we were to feel such gravitational waves, what would they feel like?

I'll start here as we need to have some well established meaning for how we can feel a wave before calculating what conditions produce such a wave. However, this is the ambiguous part. A gravitational wave passing over us would effectively just be felt as a differential gravitational field, which can induce spaghettificiation as Andy points out. One side of your body might feel a stronger pull than the other being that one side is experiencing slightly different gravity than the other. But I have to say that you'd need extreme waves to have this effect. The frequency of the waves would have to be extremely high for the wavelength to be on the order of the human body's size. If the wavelengths are much longer than your body, the change, even from a powerful gravitational wave, would probably be too slow for you to detect.

I'll note that the frequency of the waves detected by LIGO were about $250\:Hz$ for the first detection (at the peak) and $450\:Hz$ for the second detection (again at the peak). That implies a wavelength of $\sim600\:km$ at the shortest. At these extreme wavelengths, the gravity differential is negligible. You'd need wavelengths on the order of a meter which implies frequencies of $10^9-10^{10}\:Hz$ which is right near the peak of astrophysical possibilities. So already your proposal of Earth-sized black holes is out. There's no way they can produce these extreme waves.

But now we run into another problem. In order to make the wavelengths on the order of the human body such that you can actually feel the differential, you've made the frequency extremely high. Could your body even notice gravitational changes happening in $10^{-10}\:sec$? I doubt it. That'd be like me blasting you with hot air, then cold, but flipping back and forth between the two at a rate of 10 billion times a second. I have a feeling everything would just cancel out and you wouldn't really feel anything, but it's hard to say.

  How close would a black hole collision have to be for us to feel the gravitational waves pass through Earth?

Let's ignore the issue of whether you'd ever feel a wave and look at the main question. I'll just consider raw energy in the wave and hope that somehow equates to feeling it. The power output of two orbiting bodies of mass $M$ (assuming the same mass for simplicity), and orbiting at a radius of $r$ is given by

$$P_{grav} = \frac{32}{5}\frac{G^4}{c^5}\frac{2M^5}{r^5}$$

Note that $G=6.67\times10^{-11}\:m^3kg^{-1}s^{-2}$ is the gravitational constant and $c=3\times10^8\:m/s$ is the speed of light. Now $P_{grav}$ is the total power radiated out in energy per second in the form of gravitational waves. You of course would not be experiencing this entire power though because this is radiated in all directions. The power is not radiated spherically symmetrically, but for simplicity's sake, let's just assume it is. That means we can say the power you experience given your area $A$ and distance $d$ from the system is given by

$$P_{exp} = P_{grav}\frac{A}{4\pi d^2} = \frac{32}{5}\frac{G^4}{c^5}\frac{2M^5}{r^5}\frac{A}{4\pi d^2}$$

Play with some numbers for this and see what is necessary to have an appreciable power.

I will point out though that at the distance you'd have to be where this power became appreciable, it would be overwhelmed by the tidal forces involved.

So in short, no I don't think you'll ever be able to feel gravitational waves, no matter the circumstance.


How could the Moon have arisen from a collision - what would make for an elliptical orbit? One serious theory advanced for the creation of Earth's moon is that it resulted from a collision.

However matter would be thrown off in all directions and some will fall back. How could a mass of ejecta as large as the Moon all end up travelling together in a single direction and therefore a single orbit? For a spacecraft to attain orbit requires some form of propulsion. Even if the impacting object 'bounced off' I'm still not clear that it could end up where the Moon is and travelling in the right direction. This would require it to be 238,900 miles away and travelling roughly at right angles to the Earth's gravitational field.

  The giant-impact hypothesis, sometimes called the Big Splash, or the
  Theia Impact suggests that the Moon formed out of the debris left over
  from a collision between Earth and an astronomical body the size of
  Mars

  https://en.wikipedia.org/wiki/Orbit_of_the_Moon

How could a mass of ejecta as large as the Moon all end up travelling together in a single direction and therefore a single orbit?

The Moon did not form this way. The giant impact hypothesis instead proposes a collision between the Earth and an object that was significantly larger than the Moon, roughly ten times the mass of the Moon. Per this hypothesis, large amounts of material ejected from the surface of the Earth. The energy released from the collision would have heated the impactor and the surface of the Earth immensely. The proto-Earth would have been enshrouded in an atmosphere of vaporized rock. Meanwhile, a handful of Earth radii out, there was an orbiting collection of bits, a few somewhat large, but most rather small.

Over the course of decades to centuries, these bits coalesced to form what would become our Moon. The Earth cooled and formed oceans. The close-by Moon raised huge tides on the Earth. Like today, but amped up by orders of magnitude by those huge tides, the tidal interactions between the Earth and the Moon made the Moon's orbit recess from the Earth. Now we see our Moon at about sixty Earth radii from the Earth, compared to the roughly one tenth of that when the Moon first formed.


Could you use luminosity changes over time to calculate distance? More specifically: if you know your movement relative to a body, could you use the inverse square law to calculate distance? Since if you are distance $D$ away and you move $d$, $\Delta L \propto (D-d)^2-D^2=2dD-d^2$, I'm assuming you should be able to calculate D linearly.

It is an interesting question, but it won't work because the luminosity of astrophysical sources is not constant enough, and because measurements cannot be taken with the required precision on long timescales.

Let's assume that you have an astrophysical sources that has a constant luminosity (note, I think you have confused luminosity with apparent brightness) and which has a perfectly measured line of sight speed $v$.

At some particular moment
$$  F(t) = \frac{L}{4\pi D(t)^2}\ ,\tag*{(1)}$$
where $L$ is the luminosity, $D(t)$ is the (changing) distance and $F(t)$ is the observed flux.

We could observe how the brightness changed with time
$$ \frac{dF}{dt} = \dot{F} = -\frac{2L}{4\pi D^3}v\ . \tag*{(2)}$$

If the luminosity is unknown, we can eliminate it by combining equations (1) and (2) and rearranging to give $D$ in term of the observables, $F$, $\dot{F}$ and $v$.
$$ D = -\frac{2F}{\dot{F}}v\ , \tag*{(3)}$$
where a positive $v$ indicates movement away from us.

So it could work in principle, but let's put in some numbers. Let's say we have a source at $D=10$ pc and let's give it a big stellar velocity, say $v = 100$ km/s. That would mean (from equation 3), that $\dot{F}/F = 6\times 10^{-13}$ s$^{-1}$ or
$$\frac{\dot{F}}{F} \sim 2\times 10^{-5} \left(\frac{D}{10\ {\rm pc}}\right)^{-1} \left(\frac{v}{100\ {\rm km/s}}\right)\ {\rm year}^{-1}$$.

So you would have to have instrumentation capable of measuring a change in brightness of 2 parts in 100,000 every year, assuming that the luminosity didn't change at all. And the required precision gets more demanding for larger distances or for slower velocities.

The Sun's (a normal star) brightness varies by 1-2 orders of magnitude more than this every year and there just isn't any astronomical instrumentation that has the capability or stability to make such measurements. Therefore it is unfeasible for at least two reasons.


How do we know that our galaxy is a spiral galaxy? I know that our galaxy is spiral in shape, but I'm wondering how the scientists found out that our galaxy has a spiral shape.

I don't think we can see the entire galaxy from telescopes on Earth, right?

I think it makes sense that they say that Andromeda has a spiral shape because we can see the whole galaxy, but how do scientist know about our galaxy?

To elaborate on Conrad Turner's excellent list:

The disk structure is even more easily seen in the infrared, where dust extinction is much less than in the optical. See for instance this image from the 2MASS infrared survey:

Stars are not really (as Milan van Dijck suggests) used for mapping the spiral arms. Rather the gas clouds of gas in between stars are used, more specifically neutral hydrogen observed through its  21 cm observation. The figure below (from Oort 1959) shows the resulting map. As you can see, its not really as well-defined as seen in many artist's-conception pictures.

Neutral hydrogen is fuel for stars. Massive stars ionize the gas around them, and since star formation mainly takes place in the spiral arms, the gas clouds are accompanied by pockets of ionized hydrogen, as seen in this map from Georgelin & Georgelin (1976):

The physical properties of the Milky Way (stellar population, gas fraction, color, kinematics, etc.) are also similar to other spirals.


Is the boulder on the peak of Tycho Crater the core of the impactor, or is it a random rock? There's a strikingly visible boulder sitting right at the top of the peak complex in the center of Tycho Crater on the Moon.
Is it just a coincidence that this gigantic, bright rock is sitting dead center of the peak complex or is it a known piece of the original impactor that caused the wider crater? Does this remarkable rock have a name?
Image source

On 10 June 2011 the LRO spacecraft slewed 65° to the west, allowing the LROC NACs to capture this dramatic sunrise view of Tycho crater. A very popular target with amateur astronomers, Tycho is located at 43.37°S, 348.68°E, and is ~82 km (51 miles) in diameter. The summit of the central peak is 2 km (6562 ft) above the crater floor, and the crater floor is about 4700 m (15,420 ft) below the rim. Many "clasts" ranging in size from 10 meters to 100s of meters are exposed in the central peak slopes. Were these distinctive outcrops formed as a result of crushing and deformation of the target rock as the peak grew? Or do they represent preexisting rock layers that were brought intact to the surface? Imagine future geologists carefully making their way across these steep slopes, sampling a diversity of rocks brought up from depth. Tycho's features are so steep and sharp because the crater is young by lunar standards, only about 110 million years old. Over time, micrometeorites, and not so micro meteorites, will grind and erode these steep slopes into smooth mountains. For a preview of what Tycho's central peak may look like in a few billion years, visit Bhabha crater.
This is a version with about one fourth the linear pixel density of the original NASA full size image.

Original sources 1, 2  Credit: NASA / GSFC / Arizona State University

It is just a rock.
A complex crater like Tycho is formed in several stages as the rock behaves like a fluid. The initial impact completely destroys the impactor and excavates a large cavity in the moon, and ejecta is shot out to all sides.
Powerful shock waves push the rock to the side building large walls. As these push down on the edges of the crater, the centre is pushed up forming a central peak. Further faulting and cracking will continue to modify the crater and the peak.
It is possible that this is part of the breccia that falls back to the surface having been shot upwards in the initial impact, but that seems very unlikely. It is more probable that this is part of the ridges of the central peak that cracked and fell into the shallow valley on the top of the central peak, which formed during the faulting as the peak was pushed up from the crater floor. In all likelyhood, this is a piece of moon rock. As it hasn't been exposed to solar radiation for billions of years, it is comparatively light. It can't be a piece of the impactor, as that would have been completely destroyed


Do/did the asteroids contain enough water to create Earth's oceans? It seems from the recent Science paper by Altwegg et al (2014) that the commonly accepted source of Earth's water being the comets might not be (completely) true. Their study suggest that the water in the comet they studied is different from Earth's (higher Deuterium content). The alternative source of water that Altwegg suggests is from asteroids. My perception is that there is not that much water in asteroids or at least not sufficient to create Earth's oceans. How much water is in asteroids now? How much water was there in asteroids reaching Earth when the oceans were formed?

This is far too soon to draw sweeping conclusions. The Philae probe has only measured deuterium for one comet. But comets are diverse, of diverse origins. We need to take a lot more samples, from many different comets, before we can conclude firmly that water has, or has not, come from comets.

The only conclusion we could draw so far from this study is that earthly water has not come exclusively from comets similar to 67P/C-G.

What you're seeing now is the usual distortion that occurs in the media after some hugely popular experiment is completed - its conclusions are blown out of proportion. But that's not how science works.

Let's write this one down, and wait for further science to be put forth. We don't know enough yet about these objects.


Accuracy of Mercury transit calculations What is the error range for calculations for the time of 1st-4th contacts for a given GPS coordinate on Earth?

I observed through a telescope the 3rd and 4th contacts and compared against SkySafari and Stellarium.  I found SkySafari showed contacts about 10 seconds earlier for the same location vs Stellarium.  Stellarium was very close to my measurements for 3rd contact (1-2 seconds).  To get the calculation I zoomed the GUI and tweaked the time until it was closest to the contact.  Estimates for 4th contact (transit end) are much harder to visually time.

This got me interested to know if there is a gold standard for such calculations, and what is the range of error for it.   For forecasts n years out is the error range n times more?

This got me interested to know if there is a gold standard for such calculations

For the orbital mechanics and geometrical aspects, the NASA JPL Development Ephemerides are commonly used. These have errors for the Sun and inner planets of tens of kilometers or less.

How to pronounce “Ephemerides”?

  ...and what is the range of error for it. For forecasts n years out is the error range n times more?

That depends on the specific ephemeris you use and a lot of other details. The ephemerides are build from a huge body of observations and calculations and it's not a simple thing to say. I think the errors will be dominated by the definition of the edge of the Sun for several decades or even centuries. Farther out in time, the ephemeris itself will have some error that may increase with time in either direction.

Below I calculate that the apparent radial speed at which mercury crosses the "edge" of the Sun's disk is about -70.6 and +70.6 km/sec, assuming a "radius" of 695,700 km, using the Python package Skyfield which in turn uses the NASA JPL Development Ephemerides. For this calculation I used DE421.

I used Caracas as the location since the whole event was visible from this location. The main calculation happens in these three lines. The .observe() method automatically does the light-time correction for the positions.

sun_astrometric     = caracas.at(times).observe(sun)
mercury_astrometric = caracas.at(times).observe(mercury)
angular_separation  = sun_astrometric.separation_from(mercury_astrometric).radians

I put the words "edge" and "radius" in quotes earlier because these are not well defined for things like stars and gas giant planets. There are definitions that can be used to define nominals for these, in the case of the Sun it's when the opacity reaches 2/3 or something like that.

70 km is a pretty small distance at the edge of the Sun. From this answer you can see that there is a region of a few hundred kilometers where things are changing. It's hard to say exactly what defines the "edge" of the Sun in order to say when the hard edge of rocky Mercury will be seen to touch it. I can't be sure, but I have a hunch that you can get several seconds of ambiguity based on how the edge of the Sun is defined.

Wikipedia even lists **two different values for the radius that differ by 642 km*!* At 70 km/sec that's almost ten seconds right there!

Equatorial radius
  695,700 km  [7]
  696,342 km  [8]

From this answer:

  From Chapter 2: The Photosphere of Timo Nieminen's thesis Solar Line Asymmetries: Modelling the Effect of Granulation on the Solar Spectrum

    Figure 2-3: The Holweger-Müller Model Atmosphere

    7 Holweger, H. and Müller, E. A. “The Photospheric Barium Spectrum: Solar Abundance and Collision Broadening of Ba II Lines by Hydrogen”, Solar Physics 39, pg 19-30 (1974). Extra points have been cubic spline interpolated by J. E. Ross. The optical properties (such as the optical depth and the opacity) of a model atmosphere are, obviously, very important, and will be considered later. See table C-4 for complete details of the Holweger-Müller model atmosphere including all depth points used.

    8The height scale is not arbitrary. The base of the photosphere (height = 0 km) is chosen to be at standard optical depth of one (i.e.  5000Å = 1 ).

Python script using the Skyfield package to calculate some things about the 11-Nov-2019 transit of Mercury across the Sun as seen from the city of Caracas, Venezuela.

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

# https://astronomy.stackexchange.com/questions/33918/accuracy-of-mercury-transit-calculations

load        = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files
data        = load('de421.bsp')
ts          = load.timescale()
bodies      = ('sun', 'mercury', 'earth')

seconds               = 12*3600. + np.arange(0, 390*60+1, 10)
hours                 = seconds/3600.
times                 = ts.utc(2019, 11, 11, hours)

sun,  mercury, earth  = [data[x] for x in bodies]
caracas               = earth + Topos(latitude_degrees  = 10.48,
                                      longitude_degrees = -66.90,
                                      elevation_m = 900.)

sun_astrometric     = caracas.at(times).observe(sun)
mercury_astrometric = caracas.at(times).observe(mercury)
angular_separation  = sun_astrometric.separation_from(mercury_astrometric).radians
distance_to_sun     = sun_astrometric.distance().km
distance            = distance_to_sun * np.tan(angular_separation)
r_sun               = 695700*np.ones_like(hours)
thing               = distance < r_sun
i_in                = np.argmax(thing)
thing[:i_in+1]      = True
i_out               = np.argmax(~thing)
hours_in, hours_out = hours[[i_in, i_out]]

dt                  = seconds[1] - seconds[0]
ddistance_dt        = (distance[1:] - distance[:-1])/dt

if True:
    plt.figure()
    fs = 12
    plt.subplot(2, 1, 1)
    plt.plot(hours, distance)
    plt.plot(hours[[i_in, i_out]], distance[[i_in, i_out]], 'ok')
    plt.plot(hours, r_sun, '-k')
    plt.ylim(0, None)
    plt.title('projected distance from Sun center (km,)', fontsize=fs)
    plt.subplot(2, 1, 2)
    plt.plot(hours[:-1], ddistance_dt)
    plt.plot(hours[[i_in, i_out]], ddistance_dt[[i_in, i_out]], 'ok')
    plt.xlabel('hours UTC 2019-11-11', fontsize=fs)
    plt.title('d projected distance/dt (km/s)', fontsize=fs)
    print(ddistance_dt[[i_in, i_out]])
    plt.show()


Did cosmological inflation occur at speeds greater than $c$? If cosmological inflation occurred at speeds less than $c$, wouldn't we see the CMB right in front of our noses (or not at all) instead of at 13.7 billion light years?
And if it happened faster than $c$, does that mean inflation occurred before the current laws of physics (read the four basic forces) precipitated, was $c$ greater then than it is now?

The simple answer to your question is "yes" - the universe expanded at much greater speeds than $c$ during the inflationary epoch. This period of time was very quick but very dramatic, lasting from about $10^{-36}$ to $10^{-32}$ seconds. The universe expanded, in this very short period, by a factor of $10^{26}$. That's pretty incredible, when you think about it.

Inflation was originally proposed to solve, among other things, the horizon problem - that is, why the universe is isotropic and homogenous (on a large scale). This would mean that all the parts of the universe were in "causal contact" at one point in time. Inflation is the explanation for this.

Now, what does this all have to do with the CMB? Well, the temperature of the CMB is the same throughout the universe - a toasty 2.7 Kelvin. For the temperature to be uniform, all the regions of the universe would have had to be in causal contact in some point in time; hence, inflation explains the uniform temperature of the CMB.

However, the CMB was not around during the inflationary epoch. Far from it. It formed a lot later, when the universe was at the ripe old age of $379,000$ years. But the reason that it was formed (photon decoupling) equally throughout the universe is because the conditions were roughly equal, because of inflation. The CMB was, and still is, everywhere. It was never around during inflation, and as such was not otherwise affected by it.

I hope this helps.

My sources for the times:

Inflationary epoch

NASA


Are we less safe now that Arecibo is no more? update 3: uhoh!

Source: Reddit
update 2: Sadly, parts of the dish have collapsed, several cables have snapped, and the Arecibo radio telescope is now decommissioned and demolition has been recommended because it is now too unsafe to do anything else.

NSF begins planning for decommissioning of Arecibo Observatory’s 305-meter telescope due to safety concerns
Iconic radio telescope in Puerto Rico to be demolished

related in Space SE: How many total hours had Arecibo transmitted a signal?

update 1: Science November 9, 2020: Second cable breaks at Puerto Rico’s Arecibo telescope

The NASA News item Arecibo Observatory Returns from Tropical Storm Isaias Lockdown to Track Asteroid for NASA begins:

The National Science Foundation facility, managed by the University of Central Florida, determines that a newly discovered asteroid won’t threaten Earth

Of course right after this there was a serious failure, a support cable for the focal plane instrumentation broke and seriously damaged the dish as well.
Are we currently at elevated risk to getting hit by an asteroid now that Arecibo can not help in making precise range-rate measurements of asteroids? Optical tracking provides detection and some orbit reconstruction, but the precise range and radial velocity measurements that can be done by radar can provide very precise data that can help distinguish an impact trajectory from one that will miss.
Could the Five-hundred-meter Aperture Spherical Telescope fill in for Arecibo's role in asteroid astrometry?

The linked article is copied from a university press release.
Arecibo's article is more matter-of-fact but naturally also emphasizes the value of their own work.
NASA Goldstone can do planetary radar and remains operational.
As this page says:

Arecibo has twice the range and can see three times the volume of Goldstone,
while Goldstone, whose greater steerability provides twice the sky coverage and much longer tracking times, serves a complementary role.

FAST has no transmitter but presumably could take part in bistatic radar observations as GBT does.
Radar can measure distance and radial velocity ($r$, $\dot{r}$) in a single observation, especially useful when combined with astrometry ($\alpha$, $\delta$) since the two different kinds of data are orthogonal.
Given enough observations over a long enough time, astronomers can determine asteroid orbits rather precisely through optical astrometry alone.


Is the Sun really a medium size star? It is often said that the Sun is a medium size, or even dim star. Is this true?

According to this list of stars within 21 light years, there are, out of the closest 121 stars, only six brighter than the Sun. This means that the Sun is in the top 6%. If you also count brown dwarfs, the Sun ranks even higher.

Doesn't this indicate that the Sun is actually a relatively large star?

It is true that a surprisingly large number of stars are smaller (and thus less massive) than the Sun. However, the stars that are bigger than the Sun are often much bigger.

Look at this chart:

Image courtesy of Wikipedia user Jcpag2012 under the Creative Commons Attribution-Share Alike 3.0 Unported license.

Notice how small the Sun is compared to some of the other stars. It's tiny! It is indeed a small star - in technical terms a main sequence dwarf.

However, despite its size, it is clear that there are many more stars less massive than the Sun that there are stars more massive than the Sun. Why? There are two reasons:

Lower-mass stars live longer.
More low-mass stars can form in a given region than high-mass stars.

Encyclopedia of Astronomy and Astrophysics

The distribution of masses can be quantified in an initial mass function, typically given in the form
$$\xi(m)=km^{-\alpha}$$

When you integrate this over a range of masses, you can find how many stars are within that range. Not surprisingly, this number gets lower and lower as you slide the endpoints to more massive stars. You can see this decrease from the fact that $\xi'(m)<0$, so long as $k>0$ and $\alpha>0$ - which is assumed by the model, according to empirical data.


How to listen to our world from 10,000 light years away? If you were 10,000 light years away from all radio stations and digital broadcasts from our world, what equipment would be necessary to detect and capture some information like a voice or a photo? If we found a similar planet in the galaxy, what could we do to listen to their radios?

As the above commentator says, our technology isn't sufficiently sensitive to pick up everyday radio broadcasts from so far away, and even if we could it wouldn't be worth sending them a message. The earliest we could hope to receive a reply would be in 20,000 years time, by which time there might be no humans left on Earth to receive it, as a result of nuclear war or some other catastrophe. If this extra-terrestrial civilisation were much, much nearer, tens of light years rathe than thousands, prospects would be very much better, but it is extremely unlikely that an advanced alien civilisation is as close as that.


Could a distant black hole (detected due to its mass) be in reality a Dyson sphere? When the black hole is only detected due to its gravitational effects on other stars and their light, could it be possible that it is in reality a working Dyson sphere?

It would still have at least the mass of the star but also blocks all the emitting light.

I got the idea from watching this Kurzgesagt video about Dyson spheres

The thing here is that a Dyson Sphere doesn't "block all the emitting light" -- that would violate the conservation of energy.  It does absorb the radiation from whatever body it surrounds and then re-radiates it.

As such, it's likely to radiate something like a relatively normal-looking blackbody spectrum, rather than the plasma X-rays from matter falling into a black hole, or the almost-nothing (thus "black") from a black hole with no infall.

As pointed out in @SteveLinton's comment the Dyson sphere would be much larger than the star or a black hole of the same mass in order to not melt, so at that distance the deflection would be far less, since it scales as inverse radius:

$$\approx \frac{m}{r}$$.


Could a moon orbit an Earth size planet in such a way as to create a solar eclipse lasting for several weeks from start to finish? Is it theoretically possible for a  moon to orbit a planet (let us say Earth size, mass and solar distance) in such a way as to create a planet-wide solar eclipse lasting several weeks?

These are my assumptions (some of them likely flawed) about the circumstances under which this might be possible:

To simplify the problem, let us assume that there is no planetary axial tilt and that the moon has an equatorial orbit.
The orbital speed of the moon would need to be quite slow in order to stay in position for long enough to create a graduating eclipse over several weeks.
In order to create a planet wide eclipse, the moon would need to be quite large and/or relatively close to the planet. Is this circumstance feasible if the mass of the object were low compared to its size?
How stable could the described relationship be?

Thank you for your interest and answers.

Short answer: No, it could not.

The moon already has little enough mass relative to the earth so that its orbital period is not much shorter than a satellite with negligible mass. So, any object at about the distance of the moon would take around a month. If the object really had negligible mass it would be moving a little slower than the moon, so let's be generous and say 5 weeks. That means the object would have to be large enough to extend 1/5 of the way around the sky to block the sun for just one week. This is not possible. Such an object would be much larger than the earth.

If you imagine an object closer to the earth, it would be orbiting faster so it would have to be even larger.

Moving farther away is a little better. The orbital period is proportional to the radius raised to the 3/2 power, so moving the moon four times farther away would make its period eight times as long. But it would have to be four times larger, or 64 times the volume, just to eclipse the sun at all. And at that point it's approaching the size of the earth.

It may be possible to construct a synthetic object that orbits the sun at earth's L1 point (between the earth and the sun at just the right distance to stay in line with them) and permanently blocks the sun. You could tweak its solar orbit so that it only blocked the sun part of the time. It could be something like a mylar disk that rotates to avoid collapsing into a ball. But this becomes more of an engineering question than an astronomy one. No plausible natural object would have this configuration.


What is the next planned mission that can discover life on another planet? I've heard that we have a great shot at finding life on Mars if we just drill deep enough down to the surface to the water and detect if its sterile or not. Are there planned missions to do this? If not, what is the next (or existing) mission where finding life is likely? Is it launching something to Mars, Europa, Titan, or some other moon? Why does the mission have a good shot at finding life?

The Exomars rover of ESA and Roscosmos would be the obvious answer. To be launched in 2018. It will drill 2 meters deep and as far I know is the only mission since the Vikings in the 1970's, to explicitly be equipped to find biosignatures, signs of life. But the Russians, who will land the rover, have had a very poor Mars mission success rate, ESA cooperates with them since NASA abandoned the project a few years ago, and ESA's only own landing attempt on Mars failed too, Beagle 2. And some biologists think it is more challenging to detect sparse exotic microbial life than what that rover is capable of. It weights about 1/3 of MSL Curiosity, and underground life can maybe be very local.

I want to recommend the blogger Robert Walker who writes at great length and well informed, still interestingly speculative, about possibilities for life on Mars, what one maybe should be looking for.

I should add that since 1960 SETI uses telescopes to pick up evidence of interstellar life which is powerful enough to somehow change its environment to make itself astronomically detectable. That's the other main potential except for probes in the Solar system. But not even SETI people sit up waiting for it any more.


How did space start? How did this 'black fabric' that we call space start? How was it made? How was the initial singularity formed? How did the first atom come into existence?

If it all started from a tiny particle that expanded and is now still expanding, how did that tiny particle come into existence ?

I'd like to add onto what James K said, but I'll try and simplify things too.

A singularity may sound like everything was at a "single point", but that's not what the word means in physics/math. A singularity, loosely speaking, is just a point where a function reaches infinity. It's not a physical object, but an input to a function. As James mentioned, 1/x has a singularity at x=0. At the moment of the Big Bang, several functions have singularities, as I discuss below.

So with that set aside, let's explain the Big Bang. Our modern theory of gravity, general relativity, indicates that the universe is currently expanding (i.e. space is being created between all matter). By that logic, we expect the matter in the universe to be closer together as we look further back in time. Our physical theories indicate that at earlier periods, temperatures were so high that even atoms couldn't form.

Our theories can handle these conditions pretty well, but eventually they start to fall apart. At a point around 13.8 billion years in the past, our calculations predict that the amount of space in the Universe should be zero. As a result, density, gravity and temperature would skyrocket to infinity — in other words, they have singularities at that moment. This moment is known as the Big Bang.

These infinities are a problem, since our theories break down at such extreme conditions. The Big Bang is an unsolved problem that modern physics has yet to explain. It's not the beginning of the universe per se, but rather we do not know what happened around, or before, that moment.


Home telescope for viewing 2017 VR12 News has come up lately about the impending approach of asteroid 2017 VR12 and how it might be visible to some backyard telescopes. However I've read conflicting or vague reports on what kinds of telescopes will be powerful enough to spot it.

I've got a modest but still decent little Celestron NexStar 4 SE, which is a 4" Schmidt-Cassegrain. Can anyone say for sure whether that has the power to see the asteroid? Currently I just have the stock eyepiece. Would I need to get a better one to help out?

2017 VR12
is expected to reach magnitude 11.8 on the night of March 6-7.
This is near the limit of a 4-inch scope; actual visibility depends on your observing site, your eyes, and the magnification.
You can get a more definite answer by trying to see stars of similar magnitude.

The other hard part is to acquire the target.
It will look like a star, so you need a finder chart showing stars at least as faint as the asteroid.
If you haven't observed one before, try finding a
named asteroid
around magnitude 10 first.
In this close approach, 2017 VR12 will move 0.9 degree per hour north-south,
fast enough that you need either quarter-hour marks plotted on the finder chart or a continuously updated display.

Stellarium
works pretty well for me; I downloaded some of the extra star catalogs, and I import the latest available orbital elements for the asteroid of interest.
If you wish to use your scope's go-to capability,
this guide
may help, but you still have to figure out which "star" in the eyepiece field is the asteroid.
This article
identifies a few brighter stars you can try to watch it pass.

I do recommend getting a ~10mm eyepiece in addition to the 25mm one you have.
It might help you see stars half a magnitude fainter, but the main benefit is higher magnification for planets.


How might Thales have predicted a solar eclipse? Thales of Miletus purportedly predicted a solar eclipse (most likely the solar eclipse of May 28, 585 BCE). The historicity of Thales' prediction can't be verified with any certainty given the sparsity of written records documenting the pre-Socratic era that have survived to present day. So, putting the historical question of whether or not he did aside, what plausible answers are there to the question of how he could/would have made such a prediction?

I recently read the relevant parts of the Cambridge Concise History of Astronomy and I'd suggest it was partly good observation, partly luck.

Good observation in that Thales would have noted that the movements of the Sun and Moon were such that eclipses were possible on certain dates, and luck in that the shadow of the eclipse he suggested was possible passed over him or those to whom he had suggested the eclipse took place.

I very much doubt he had the observational data (about the size of the Earth for instance) to have been precise about the shadow path - but would have had the data to show when the Sun and Moon were in alignment.


Could an asteroid knock the moon out of its orbit? As describe in this CNN article, an asteroid is going to pass near the Earth on Halloween.

The article says that

  As it misses Earth by about 300,000 miles (slightly farther away than the moon), the asteroid, named 2015 TB145, will be visible to those with good telescopes

and that

  It's a good thing it will miss, though. The asteroid is estimated to be 300 to 600 meters wide and traveling at 78,000 mph.

If, hypothetically, this asteroid actually hit the moon, could it knock it out of orbit? How about another, contrived asteroid with whatever properties are needed?

To be clear, when I say "out of its orbit" I'm including both the orbit decaying and the moon crashing into Earth, and the moon being knocked out into space away from the Earth's influence. Either would do.

Basically no and not by a long shot.   The asteroid's measurements are pretty vague, but lets give it a nice clean 600 meters in diameter and the same density of the moon.   The moon is 3,474 km in diameter or 3,474,000 meters.   So the moon is nearly 5,790 times the diameter and that gives is 194 billion times the mass as that asteroid.    Conservation of momentum, if the asteroid transfers all of it's 78,000 mph into the moon, with no energy going into angular momentum, 78,000 / 194,000,000,000 = .0000004 mph.   That's roughly the speed your fingernails grow.

To knock the moon out of orbit, it's velocity would need to be increased by oh, very rough ballpark, maybe 400-500 mph.   The additional velocity needed for the moon to reach escape velocity from the Earth is .414 x 2288 or about 947 mph, but given that it has an elongated orbit and that it would escape the Earth's sphere of influence well below escape velocity, maybe as little 400 extra MPH could knock the moon into a far enough orbit that it could escape the Earth.

Figuring 78,000 MPH, and you want to move the moon some 400 or more MPH with ah it just right, you'd need an object about a 5th or 6th the diameter of the moon or 600 KM in diameter give or take, assuming it was straight on impact as a glancing blow would lose a share of energy into angular momentum and assuming it hits the moon away from the Earth, not towards as towards would require even more energy.

This space pebble doesn't even come close.   You'd need something the size of a small dwarf planet to have a chance at knocking the moon out of the Earth's orbit and that speed.

300-600 meters is huge, if it hits the Earth.  That's probably a few square miles of earth practically leveled to the ground and broken windows and knocked down trees a good deal further than that.   If it hits the moon it would leave a nice crater, maybe a miles or two across but that's about it.

  It seems like you're only talking about the moon being kicked out into
  space. Would the same amount of energy be needed to knock the moon out
  of its orbit in towards Earth?

The mathematics of crashing is a bit more complicated.   Escape Velocity vs Orbital Velocity is easy, just straight multiplication.   More details here, and because the Moon wouldn't need to reach full escaple velocity, I cut the number in half - and it might be a bit less than that.

Moon's orbital Speed (2,288 MPH) x (Square root of 2 minus 1, or 0.414).

Crashing the Moon into the Earth - a bit harder.  There's probably a simple enough formula, but I'm not sure what it is, but I believe that would take even more energy.

It's discussed here without math and here, with math.  But it would take a huge amount of energy, probably 3 or 4 times more energy to push the Moon into the Earth than to push it out of orbit if I was to guess.


Could the big bang have created super massive black holes? I understand that space was compressed to a single point and that during the big bang all points within that expanded away from each other at phenomenal speeds. I also have heard that during this the universe wasn't the same everywhere, as in small quantum fluctuations that expanded to macroscopic scales.

Would this mean that higher density than normal areas could create massive black holes? Maybe not as big as the ones today but perhaps large enough to solve some of the problems with super massive black holes being observed in a short amount of time after the big bang

I am not asking "why didn't the universe turn into a singularity?" I am asking "Could the variable density afterwards create many singularities which could explain for the problems associated with super massive blackholes?".

Arxiv prodives Haiman 2012 paper The Formation of the First Massive Black Holes. The leading theory, based on WMAP, is that the first perturbations resulted in dark matter halos, which were the first gravitational concentrations of matter.

  the first nonlinear objects in the universe were born inside ∼ 10^5 M⊙
  dark matter halos at redshifts of z ∼ 20−30

The non-linear objects do not refer precisely to black holes. The key part for us:

  [...] question of how the  first SMBHs  were  assembled. It is  worth
  emphasizing that this is an unsolved problem – indeed, it is not
  entirely clear even whether the first nonlinear objects in the
  universe were stars or black holes, and whether galaxies or their
  central black holes formed first.

The article proceeds to tell a very wide range of theories, but the summary is - we are quite clear about simulating the early dark matter halos but we really don't know how to clearly show whether/how it led to SMBH formation.

I'd like to emphasize what is (maybe) obvious: SMBH requires a very concentrated gravity. We assume that a very early universe despite being dense didn't have enough gravity. All theories are generally about how density dropped far down and in what ways could it start to grow again, but now only inside the concentrations (i.e. when the gravity started to affect universe, the density was much lower that inside stars or black holes).