How to compute the perihelion precession of Earth's orbit? I want to simulate the perihelion precession of Earth. I tried searching  other posts in stack exchange as well relating this question but may be it's my fault that I couldn't understand their answers. I read the history about mathematicians and physicists trying to find a good theory accounting the precession. I read that there are many planetary equation by Newton,Laplace,Hill,Poincare etc. But I am not sure which one gives the more accurate result. I am not even sure if I going on a right track searching the correct theory.
I'd really appreciate if anyone could help me with the theory which gives fast and more accurate result?

For simulation only, you should be able to show it by simple code as following:
$$
\vec{a}_t = - \frac{GM}{r_t^2} \hat{r} \\
\vec{v}_{t+1} = \vec{v}_t + \vec{a}_t \Delta t \\
\vec{r}_{t+1} = \vec{r}_t + \vec{v}_t \Delta t
$$
where $t$ is time step, $\vec{a}$ is acceleration, $\vec{v}$ is velocity, $\vec{r}$ is position, $\hat{r}$ is a unit vector in radial direction from the origin (set the Sun at the origin), $G$ is gravitational constant, $M$ is Sun mass in this case, and $\Delta t$ is size of time step. You will have to play around with the initial values and step size to get certain orbit representing the case. To make it more real, you derive the initial values from the real parameters, if known.


Gravitational field released during supernovae During the collapse of a star at the commencement of supernovae, the intense pressures force fusion and fission to create the heavier elements and release vast amounts of energy. As this energy/mass is released - the gravitational field must also decrease in an instant, and as this "Warping of space time" is instantly released and returns to flat space (Or flatter space) - does this create or contribute to the explosion of the supernovae itself? I have always been puzzled by the concept of the collapsed outer layers "Bouncing off of an in-compressible core" as the source of the explosion. Isn't this instantaneous "Relaxation of the curvature of the space" not what we see as the explosion?

What you are missing is that the shell theorem, which says that for a spherically symmetric mass distribution (we can count the pre- and post-supernova states as this for the sake of argument), that the gravitational field at some distance $d$ from the mass is the same as if all the mass were concentrated at the centre of the distribution.

Therefore the gravitational field due to the pre-supernova star is unchanged after the supernova at a distance $d$ until some of the mass (or equivalently in General Relativity, energy) as travelled beyond a distance $d$.

After a core-collapse supernova, if the pre-supernova star was say $15 M_{\odot}$, then what might happen is that $1.4 M_{\odot}$ is left as a neutron star remnant, whilst $10^{46}$ J of energy is released - mostly in the form of neutrinos travelling at almost the speed of light and the envelope of the star expanding outwards at tens of thousands of km/s and a kinetic energy of about $10^{44}$ J.

The neutrinos carry an equivalent rest mass of a mere $0.05M_{\odot}$, so if you were on a (indestructable) planet in orbit around the supernova, then yes, after the main neutrino pulse has passed there would be a small decrease in the gravitational force felt towards the centre of the planet's orbit (not instantaneously, the neutrino pulse lasts some tens of seconds) that would result in an outward acceleration and  the orbit widening slightly.

Sometime later (the orbital radius divided by the supernova ejecta speed) the main mass loss from the supernova would pass by and this would result in a drastic decrease in the gravitational force and a drastic widening of the orbit.


Can we see individual stars in other galaxies? Just wondering if we can examine individual stars from other galaxies, or if we are pretty much stuck with the billions that are in the Milky Way?

Yes, Edwin Hubble did that for the first time in 1919. Before that time, it was thought that the galaxies we can observe were just nearby gas nebulae located inside our Milky Way. But Hubble was able to resolve the nearby galaxies like the Andromeda nebula into individual stars. By measuring the brightness of so-called Cepheid variable stars, he was able to calculate the distance to the Andromeda galaxy. Here one uses the fact that the total power radiated by the star is related to the period of the brightness oscillations, so by observing such stars you can deduce the distance to these stars and hence the distance to the galaxy. But later it was found that there were two different types of Cepheid stars and the wrong relation had been used; the distances were actually about twice as large.


H-R diagram, Double stars Can someone help me? I have a case, like, we have H-R diagram and two stars on it. One of them is a blue star on main sequence, another is a red giant. They have equal luminosity. If they are in a double star system and seen from Earh as one star, then where on the diagram should I place this star? What luminosity amd temperature will it have?

Maybe I should sum up their luminosity and then, using Stefan Boltzmann law, calculate the temperature? But does the law work for red giants? And can I sum the luminosity?

"Maybe I should sum up their luminosity and then, using Stefan Boltzmann law, calculate the temperature? But does the law work for red giants? And can I sum the luminosity?"

This answer is not about placing the object on HR diagram, but is about the characteristic temperature as a blackbody.

It seems like you have a point source which you know that it is a combination of two stars: one blue, and the other redder. You can apply two-component blackbody: $F_{\lambda,total} = C_1 \times F_{\lambda,1}(T_1) + C_2 \times F_{\lambda,2}(T_2) ; T_1 \ne T_2$ where $F_{\lambda,total}$ is the observed specific fluxes from the point source, $F_{\lambda,1}, F_{\lambda,2}$ are blackbodies with temperature $T_1, T_2$ respectively, $C$s are normalization constants.

Blackbody approximately (to the first order) works with stars including red giants as long as you include a portion of the SED where the line absorption/emission is negligible. For example, some stars have strong absorption at less than 4000 A due to metals, therefore that portion deviates significantly from the blackbody.

You can sum fluxes, since fluxes are normalized for the luminosity distance. Since luminosity $L$ is $L = F \times 4 \pi D_L^2$ where $D_L$ is the luminosity distance, the total luminosity is $L_1 + L_2 = 4 \pi D_{L,1}^2 \times F_1 + 4 \pi D_{L,2}^2 \times F_2 = C_1 \times F_1 + C_2 \times F_2$. Therefore, you can sum the luminosity when $D_{L,1} \approx D_{L,2}$.


Is a supernova's core temperature absolute zero just before collapse? I thought a supernova's core temperature was absolute zero just before it collapsed, as it has run out of all nuclear and thermal energy, but I haven't been able to confirm this online anywhere and I was wondering if anyone can confirm this for me.

No, absolutely not. The core of a core-collapse supernova is one of the hottest places in the present-day universe. The temperature as the star runs out of nuclear fuel in its core is around 6-10 billion Kelvin. As it collapses, the core gets even hotter, perhaps as high as 100 billion Kelvin for a few seconds, before neutrino cooling starts to become effective.

We know that the temperatures are getting as high as this because at least two neutrino instruments on Earth detected roughly "thermal neutrinos" with energies of $\sim 10$ MeV coinciding with SN1987A in the Large Magellanic Cloud (see for instance Mirizzi et al. 2015).


How exactly can the Hubble Space Telescope see up to 15 billion light years? I've been researching about the Hubble Space Telescope on nasa.gov, and I've read that Hubble uses a digital camera to take pictures like a cell phone. That's all I've found so far. I can't find any information on how it can see so far away into space. Is the digital camera built to see billions of light years away, or does Hubble have some technology on it that allows it to zoom into the deep space?
If this question has already been answered, please leave the link in the answers.

Hubble is a telescope.  This means it has a big mirror to gather light and focus it onto a small area, and a system of lenses and mirrors to magnify the image formed by the big mirror.
Now,  how far can you see? 100 miles? 1000?  You'd be way out!  The stars are already trillions of miles away.  Without a telescope you can see the Andromeda galaxy, which is 14 billion billion miles away.
You can see it because it is intrinsically very bright. Because it is so bright you can see it from far far away. If you had bigger eyes, you would be able to see things that are further away and dimmer.  And the Hubble telescope has a big "eye"
Hubble has one more trick: It does have a "digital camera" and it can use this to make a long exposure. This means that the light from a long period of time is allowed to accumulate on the light-sensitive chip. This means that even dimmer objects can be observed.
By a combination of a big mirror, and long exposures, Hubble is able to get a very deep view of very dim and distant objects.


What is the LSST now? Where does LSST end and Vera C. Rubin Observatory begin? Space.com's It's Official: Vera Rubin Observatory Named to Honor Dark Matter Scientist says:

"We're here today to focus on the major renaming of the facility after a pioneering astronomer, that is intimately tied to one of the key focus science areas for this project," Ralph Gaume, Director of the National Science Foundation's Division of Astronomical Sciences, said during the event (the 235th meeting of the American Astronomical Society). "I'm pleased, very pleased, beyond how much you all know and may recognize, to today officially rename the LSST observatory as the Vera C. Rubin Observatory."
Gaume's comments were met by applause from the gathered astronomers. Observatory director Steve Kahn and other team members soon donned T-shirts sporting the new name.
The Vera C. Rubin Observatory is a federal project run by the National Science Foundation and the U.S. Department of Energy. Its first 10 years of work will be dedicated entirely to a project now known as the Legacy Survey of Space and Time. "Because we know you're all in love with the four letters LSST, we figured out a way to preserve that," Gaume said.

If I understand correctly, the telescope formerly known as LSST is now the Vera C. Rubin Observatory, but the acronym has been reincarnated.
What exactly is the LSST then? The article says that it is a project known as the Legacy Survey of Space and Time, but what exactly is a "project"? Is it the title of a proposal, or a financial instrument for assigning research funds, or something for business cards and letterhead?

A photo of astronomer Vera Rubin. (Image credit: NOAO/AURA/NSF)

The LSST Project is in charge of the construction project; building the telescope, camera and observatory and data pipelines and act as the "prime contractor". This includes designing and setting up the data processing and data distribution but the Project's job is essentially "done" once things start operating and the observatory transitions into regular operations. Once the Project delivers the observatory and it's signed off as complete, then the day-to-day operations transitions over to the NSF’s National Optical-Infrared Astronomy Research Laboratory. They will manage LSST along with other ground-based assets such as the (former) NOAO telescopes and the two Gemini telescopes. They will be the ones in charge of making the survey actually happen and keep functioning.

The Project was setup by the LSST Corporation LSSTC website which represents the institution and international partners, works with the Science Collaborations and is generally in charge/supports getting the science out of the observatory.

So as per the press release, the now named Vera C. Rubin Observatory consists of the mountain top facility in Chile which contains the 8.4-m telescope, now named Simonyi Survey Telescope after an early donor who got the primary work started, the commissioning camera (ComCam) and the main 3.2 Gpix DOE LSST Camera, along with Auxiliary Telescope and its dome and the headquarters in Tucson, AZ. This will be carrying out the 10yr sky survey, now backronymed to be Legacy Survey of Space and Time (LSST) to keep the same acronym.


What methods exist to calculate the ellipticity of galaxies What methods exist to calculate the ellipticity of galaxies and what are their drawbacks? I have asked this question about ellipticity in the SDSS but I want to know about general methods for cases where I have just an image of a galaxy.

I do so far know these methods:

ellipticity from stoke parameters or the flux-weighted second moment as described here.
from adaptive moments as described here (which would allow me to do shear calibration).
fit an ellipse on a particular isophote as decribed here.

My question is whether there are further methods I am not aware of. Does there also exist a method when I fit an ellipse to all isophotes? And how would I calculate the 'average' from it (such that the spherical center does not contribute to much)?

I would like to learn about the pro and cons of each method. I am especially interessted in methods that get close to what people would estimate by eye. I would also be glad to hear about references (textbooks or papers) for the different methods.

The first two methods mentioned in the question are really similar. They both use moments of the light distribution, only the weighting is different.

Another family of methods is 2D profile fitting to the image, with more or less sophisticated luminosity profile models. A simple case would be to fit a Sérsic profile. But instead, one could make a composite model with two or more elliptical profiles (say one exponential disk and a de Vaucouleur bulge). This would allow to separate the ellipticity of the disk from the ellipticity of the bulge. A widely used software for this is GALFIT by Chien Peng.


Is there a variation of ISO 8601 for julian calendar dates? If I want to write a date in the Gregorian calendar, I can write it in the ISO 8601 format; today would be 2016-02-20. The advantage is that there are many parsers available to let computers read in this kind of date. But what if I want to write a Julian calendar date, as astronomers often do? Is there a similar widely-accepted format, and selection of parsers, to use with the Julian calendar.

While I'm at it, what if it's an ancient date where we don't know how many solar days ago it was; instead it is an astronomical phenomenon calculated in Terrestrial Time (more or less atomic time) or by dating decay of radioactive elements? Is there a standard way of writing such a date?

Astronomers use "Julian day numbers" to indicate time. The Julian day number is the number of days since Noon GMT on January 1st 4713BC (in the Julian calendar). A time is then a single real number. For example, 00:30:00.0 UT January 1, 2013 (Gregorian), is 2456293.520833 (wikipedia)

So except to specify the epoch, the Julian calendar is not in use by astronomers to specify dates. There is no international standard equivalent to iso8601 for representing dates in the Julian calendar.

Ancient dates are often expressed as years before present, where "present" is 1950. 1950 is chosen as it preceeds most of the atmospheric nuclear testing that dumped massive amounts of C14 into the atmosphere and made carbon dating only valid for years before 1950.


Orphaned star systems and intergalactic travel (in fictional context) Context:
I'm looking for some authority on a particular idea that I brought up in regards to a new update brought out for a space travel sim game, Elite Dangerous

The vessels in that game can travel between star systems, requiring fuel- they can only jump between star systems, and have a limit on how many light years they can travel in one jump.

The latest update added a vessel that can jump up to 500 light years.

Vessels can't travel continuously to a location, it has to be point to point, i.e. teleportation (sort of)

The question:

Given the context above, can it be estimated that there are enough high orbiting, or orphaned star systems between the edge of the milky way, and the edge of the Canis Major dwarf galaxy for one of these vessels to traverse a 'corridor' of 500ly (or closer) intervals of star systems between the two galaxies?

The Canis Major dwarf galaxy is about 8 kpc from the Sun, but is only 8 degrees below the Galactic plane (and further out than the Sun). So it is about 42,000 light years from the Galactic centre and about 1150 light years below the plane.

This almost within the disc of the Galaxy itself. The Galactic disc has a density that varies pseudo-exponentially in both radial distance from the centre and with height above the plane:
$$ n \propto \exp(-r/R_0)\exp(-z/Z_0),$$
where $R_0$ and $Z_0$ are appropriate numbers for the radial scale length and the height scale length respectively, and $r$ and $z$ are the radial and vertical coordinates with respect to the Galactic centre.

In this coordinate system, the position of the Sun is about $r=25,000$ light years and $z \sim 0$ light years. The appropriate numbers for the scale-lengths/heights are $R_0 \sim 10,000$ light years and $Z_0 \sim 1000$ light years.

The density of stars in the solar neighbourhood is about 0.1 pc$^{-3}$ or about $3\times 10^{-3}$ stars per cubic light year. We can now scale this number according to the equation above, to work out the approximate density in the vicinity of the Canis Major dwarf galaxy, which will be loer than the solar neighbourhood.
$$ \frac{n_{\rm CM}}{n_{\rm Sun}} \simeq \exp[(r_{\rm Sun} - r_{\rm CM})/R_0]\exp[(z_{\rm Sun}-z_{\rm CM})/Z_0],$$
from which we get $n_{\rm CM} \simeq 0.06 n_{\rm Sun} = 2\times 10^{-4}$ stars per cubic light year.

To work out the average distance between stars we take the reciprocal of the cube root,
$$n_{\rm CM}^{-1/3} = 18\ {\rm light\ years}.$$

So, if you have a range of 500 light years per "jump" you have no problem skipping all the way to the Canis Major dwarf galaxy.


Why would atmospheric absorption increase rather than decrease the antenna temperature in Penzias and Wilson's famous CMB observation? The classic paper A MEASUREMENT OF EXCESS ANTENNA TEMPERATURE AT 4080 Mc/s begins:

Measurements of the effective zenith noise temperature of the 20-foot horn-reflector antenna (Crawford, Hogg, and Hunt 1961) at the Crawford Hill Laboratory, Holmdel, New Jersey, at 4080 Mc/s have yielded a value about 3.5° K higher than expected. This excess temperature is, within the limits of our observations, isotropic, unpolarized, and free from seasonal variations (July, 1964-April, 1965). A possible explanation for the observed excess noise temperature is the one given by Dicke, Peebles, Roll, and Wilkinson (1965) in a companion letter in this issue.
The total antenna temperature measured at the zenith is 6.7° K of which 2.3° K is due to atmospheric absorption. The calculated contribution due to ohmic losses in the antenna and back-lobe response is 0.9° K.

So, why would it increase? And, what is the significance of zenith in this sentence?

Short answer.

Great question! Ask yourself what absorbs better, something that is white, or black.Then ask yourself what radiates better. The answer is the same thing. It turns out that absorptivity and emissivity are linked pretty closely.

Kirchov's Law of Thermal Radiation can be stated as:

  For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.

If you know that the atmospheric absorption at a given wavelength is 0.01, then you could estimate the emissivity to be 0.01 as well. It doesn't mean it's exactly right, but it's a great start.

If you then say the average temperature of the atmosphere in the region where absorption (and thermal radiation) at 4080 MHz is 230K, then the Noise Equivalent Power (NEP) will be 0.01 time 230 K or 2.3 K.

How can we express power in terms of a temperature?

Wait a minute, thermal radiated power varies as something like temperature to the fourth power! How can we add two temperatures together, or multiply them by 0.01?

It's because the energy of the photons we're talking about is way down on the left edge of Planck's blackbody distribution. See for example this answer to the question Why doesn't thermal radio emission from a DSN “hot dish” completely swamp the benefits of a cold LNA? Down at photon energies well below $k_B T$ we can use the Rayleigh-Jeans law where the power at a given frequency depends linearly on temperature.

Plotted on a log-log graph, you can see that the peak heights increase three times faster than the straight-line low-frequency segments do, showing the $T^3$ dependence of the spectral density peak height versus the linear behavior in the Rayleigh-Jeans regime. If you integrated the total area under the curve, you'd get the $T^4$ dependence of the total power radiated by a blackbody.

below: From New Jersey Institute of Technology's Dr. Dale Gary's nicely-written Physics 728 Radio Astronomy; Lecture #1 notes:


Did time pass more slowly in the past? 13 billion years ago, the universe was about 600 million years old according to many scientists. At that time, all matter in the universe would have to be closer, or in other words denser.

Did time pass more slowly, at that time, relative to now? I ask, because increased gravity causes time to pass more slowly.

If we'd been there, and measured the speed of light as we measure it today, would the speed of light be different?

Finally, does distant stars look magnified, given that a "tiny" universe is stretched on our entire sky?

I assume then that we'd measure time based on caesium 133 and distance not as a function of the speed of light, but as a theoretical physical "device" that would be the same physical length then and now.

Without some reference to compare to, time passing more slowly makes no sense at all. But every clock has its own proper time that measures time along its own worldline.

In cosmological models, the cosmological time is the proper time of a certain kind of ideal observer: one comoving with the Hubble flow. In other words, imagine space filled with observers for whom the universe looks as close to the same in every direction as possible (zero CMBR dipole anisotropy). The cosmological time is what's measured by their clocks.

As such, there is no time dilation... because we defined our time coordinate as that which is measured by the comoving clocks no matter how small or dense the universe is. You can see this in the standard form of the FLRW metric:
$$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\left[\frac{\mathrm{d}r^2}{1-kr^2} + r^2\,\mathrm{d}\Omega^2\right]\text{,}$$
where nothing happens to the time component no matter what the scale factor $a(t)$ is.

How could we compare the rate of clocks now with rate of clocks then? We could have the past clock emit some pattern, say a light at a specific frequency, and then later catch and measure it. If we find a different frequency, we could interpret this as evidence of time dilation. But we could also fix the clocks as our time standard and interpret the frequency shift as the result of a changing scale factor.

We could also arbitrarily rescale the time coordinate any way we like, but there's no point in doing so because it wouldn't change the time experienced by comoving clocks.

  are you saying that we've decided to look at it as an increasing scale factor where clocks are fixed, instead of time dilation - but in fact both are just different point of views?

That was phrased too poorly, so let me clarify. If you're watching a far-away supernova, the collapse and explode process will take a different amount of time in the supernova's own local rest frame than is seen by you. Taken alone, it's up to you whether you interpret this fact as time dilation of your clock relative to the supernova or a stretching of wavelengths due to cosmic expansion, or any combination of both.

However, what we have decided is to use a frame in which the homogeneity and isotropy of the universe is directly manifest, in which case we're not "looking" at a single supernova but all around us. For a general FRW spacetime, we cannot re-interpret cosmological redshift as due entirely to time dilation without spoiling those conditions.

The reason is that homogeneity and isotropy picks out a sequence of spatial hypersurfaces, i.e. a sequence of "nows", snapshots of the universe at instants of time. Then generally the distances between different galaxies are different at different times.


What will New Horizons be able to say about a magnetic field at Pluto? Does NASA's New Horizons flyby probe to Pluto have instruments to measure Pluto's (or Charon's) magnetic field, if any?

I'm a bit confused by the Wikipedia article which says that New Horizons, during its flyby of Jupiter, on one hand:

  Travelling through the planet's magnetosphere New Horizons collected
  valuable particle readings.[45] "Bubbles" of plasma which are believed
  to be formed from material ejected by the moon Io were noticed in the
  magnetotail.[47]

But then on the other hand:

  An objective to measure any magnetic field of Pluto was dropped. A
  magnetometer instrument could not be implemented within a reasonable
  mass budget and schedule, and SWAP and PEPSSI could do an indirect job
  detecting some magnetic field around Pluto.

And how true is this story which I read or heard somewhere? That when New Horizons was being built and launched, no one really believed that Pluto could have a magnetic field. But since then planetary science has advanced and it is now plausible that it might have a magnetic field, because of tidal forces between Pluto and Charon which might make both objects somewhat fluid and possibly generating magnetic fields.

I believe, I don't know if there is evidence of it however, that Pluto does not have an active core. An active core as I'm sure your aware produces the magnetic field, without a magnetic field neither Pluto nor Charon have a magnetosphere. If there was sufficient or any evidence at all to suggest otherwise, I believe it would have become a priority.

Theoretical models of Pluto are based typically on it's orbital speed, which suggest there is no field. However as you have stated we have witnessed tidal stresses producing activity, such as Io, which should be frozen like rock however is the most geological active body we know of it the solar system. One major difference, is we don't have images of neither Plutos nor Charons surfaces, if we had images of a smooth surface we could suffice the surface at least is active through tidal forces. Being all cratered and looking ~5 billion years old indicates the opposite. There just isn't any data as yet to say otherwise, and certainly wasn't when New Horizons was being designed.

As for what the craft could detect with it's instruments, the most indicative feature would be optical imagery and the dwarfs' surfaces. In essence, all of it's instruments could be used; ALICE for detecting atmospheric properties could certainly indicate a magnetic field, whether it's able to hold onto any material spewing from it's surface for example and any ionization of any atmosphere, which in turn is what PEPSSI does. The other great tool would be Ralph, mapping thermal imagery of the dwarfs. If there is indication of internal heating unexpected from gravitational compression of the core, or veins of thermal radiation could also help determine any activity in turn producing the magnetic field.

I hope this was what you were asking.


Is it just the observable universe that is expanding? Is it possible that it is just our observable part of the universe that is expanding, in the time that we exist, and other parts are both expanding and contracting at different rates and times?

Would light from a shrinking or slower/faster-expanding part of the universe reach us?

Could that be an alternative to the big-bang theory, in that the red-shift is a temporary situation local to our time and part of the universe? A "throbbing" energy that leads to the non-uniformity that we see in galaxy clusters; something like foam awash on the ocean surface.

There is no evidence to support the idea that some part of the universe (at a cosmological scale) is contracting. Obviously one could construct such a theory but there is currently no observational data to support it.

Until relatively recently there were three broad ideas about the Universe's expansion - that gravitational pull would slow the expansion but by never enough to halt it, that (a boundary case) there was exactly enough matter in the universe for expansion to halt at an infinite point in the future, or that eventually gravitational attraction would first slow and then reverse the expansion, leading to a big crunch.

In fact, though, the observational evidence suggests that the expansion of the Universe is accelerating - that there is some "dark energy" that is speeding up the expansion. This is now the generally accepted view, though arguments continue as to what this "dark energy" is.


Is it possible that the shadow of The Moon is a single dot during solar eclipse? During a solar eclipse The Moon covers The Sun on some places on the Earth. Now the focal point is a bit above the Earth so the shadow area is about 160 miles.  In history The Moon was a bit closer to the Earth but could there have been a time that the shadow area was just a dot. (probably for just some places as the Earth is not really perfect a sphere and the orbit of the Moon isn't also a perfect circle).

What you are calling a focal point is the end of the umbra, the point at which the umbra changes to the antumbra. In a total solar eclipse, that point is below rather than above the surface of the Earth. An annular eclipse occurs when that point is above the surface of the Earth. The Moon appears to be smaller than the Sun, leaving an unbroken ring of the Sun that appears to surround the Moon.

There are some solar eclipses that transition from annular to total, and from total to annular. A point on the surface of the Earth is exactly at the end of the umbra at those times of transition points in time. At this point in space and time the Moon will appear to be exactly the same size as and exactly in line with the Sun.


How could the universe be a few light years across just 1 second after the big bang? I was skimming through Neil deGrasse Tyson's book Astrophysics for People in a Hurry when I noticed he says something curious on page 26:

"By now, one second of time has passed [since the big bang]. The universe has grown a few light years across."

How could the universe have expanded to a few light years in diameter in just one second if nothing can travel than light? By definition of a light year, wouldn't take a 1+ years for the universe to get this big?

Your question hinges on the misconception that nothing can travel faster than light. No physical object can move faster than light, but plenty of non-physical things can. For example, take the straight line $y = mx + c$. If one varies $c$, the x-intercept of the line also changes. It's possible that this x-intercept moves faster than light for a suitably small $m$. The caveat is that this intercept is a mathematical point, and it cannot be used to transfer information. Relativity isn't violated.

In the case of the universe's expansion, the physical objects in the universe aren't moving faster than light, but they can appear to do so because space is expanding. Let's take Alice and Bob, separated by one meter, and let the space between them expand at some rate. There's no speed-of-light limit on how fast this rate can be. If space expands by a billion meters per second, then one second later, Alice and Bob would be separated by a billion and one meters, but they wouldn't have traveled faster than the speed of light since from their own point of view, they're both stationary.

In the same way, our best guess of the current universe is that the expansion is speeding up. A very far away object might eventually appear to recede from us faster than the speed of light. Relativity remains unviolated, although it means that we'll eventually be causally disconnected from these faraway objects (that is, we cannot affect or even observe them, and vice versa). Something of this sort - exponentially-increasing expansion - happened in the very early universe.


Would you get the same planet-discovering data for our own system at the same distance we're getting from Earth? So there's a boatload of planets being discovered around stars in our solar neighborhood.

Kepler 22, 638 ly away, was discovered to have a planet Kepler 22b, calculated to be twice the size of Earth.

The observations that make these measurements aren't high resolution images but periodic changes in luminosity (the transit method).

My question is: would a similar situation with similar telescope properties (Kepler) detect our own Earth around our Sun from a similar distance? If not, then how close would you have to be to see Earth?

Presumably there is some correspondence among: distance, luminosity/size of a star, size of a planet. Any other factors that make Kepler-22 different from us?

Note: this question asks about Alpha Centauri. I'm asking specifically about distances like Kepler-22.

There is a lot of detail in my answer to If Alpha Centauri A's solar system exactly mirrored our own, what would we be able to detect? I showed that current Doppler techniques and transit techniques would be unlikely to detect the Earth at the distance of Alpha Cen.

It does not get any easier if you are further away. Stars are point sources at 1pc or 200pc, so these spatially unresolved techniques just care what signal level is being received - further away is worse.

In the case of Doppler techniques we need an improvement in radial velocity precision and stability of a factor of a few.

With transits, Kepler was almost there. So another Kepler with a slightly bigger telescope that observed the Sun for say 5 years would do the trick (if the orbital inclination was about 90 degrees). The reason Kepler 22b was just detected is because it has twice the Earth's radius so eclipses 4 times the area and gives a deeper transit signal for a Sun-like star. It is this contrast that is the key to transit detectability.

Maybe the missing piece of information here is that there is a floor level to photometric precision that can't be bettered even if the star is close and bright. So long as you can reach this level in a time shorter than any transit, then the distance is irrelevant, all that matters is the fractional change in the brightness during transit and the number of transits you observe. The first of these grows as the square of the planet radius (for a fixed stellar radius), the second just depends on the length of your mission divided by the orbital period.

The bottom line here is that the Earth in an Earth-like orbit around a Sun-like star would not quite be detectable by Kepler in a Kepler-length mission, even if it were very close and nor if it were further away.


Can young surface of Venus be caused by early oceans instead of global resurface event? There is evidence of a global resurface event on Venus: the craters are young and evenly distributed.

Is it possible that there is another reason: Venus had a very deep ocean in the past, and that hindered formation of craters on the young planet

Venus is believed to have had a large amount of water when it formed, but the water molecule has been broken down due to lacking magnetic field for protection, and the hydrogen released has escaped from the planet.

My hypothesis is:

In ancient times, Venus had a very deep ocean
The very deep ocean prevents formation of impact craters on the ancient planet, so no old impact craters are found on the surface of Venus
Later, the water in the ocean was broken down into hydrogen and oxygen  by solar activity, so the water level decreased.
When most water had escaped from planet, asteroids can strike the crust directly and form impact craters.
As most impact craters can only be formed after the water escaped, so we can only find young craters at the surface of Venus.

Is my hypothesis possible?

Our best guess is that if Venus had an ocean, it did 4 billion years ago. Then it had the runaway greenhouse effect due to its thick atmosphere. The runaway greenhouse effect would have boiled away the oceans if the planet ever had oceans.

There are several guesses about why its surface is so young (300 million to 600 million years old). One is that the planet get so hot the entire surface melts. It has no plate tectonics and it has a super thick atmosphere to trap the heat. All that heat melts the surface every half billion years or so. Another reason is that it has many very large volcanoes. (167 volcanoes over 100 km across.) That's far more very large volcanoes than Earth has. Those volcanoes resurface the planet and keep it young.


How did scientists determine an estimate of the number of planets greater than Sedna's size to exist in the Inner Oort Cloud? In the article, New Dwarf Planet Found at Solar System's Edge, Hints at Possible Faraway 'Planet X' (Wall, 2014) where they report on research and observations performed by astronomers that discovered the dwarf planet 2012 VP113 which is smaller than, but has an orbit severely elongated like that observed for Sedna.

The article reports the claim from the astronomers who discovered 2012 VP113 that

  Based on the amount of sky the scientists searched, Trujillo and Sheppard estimate that about 900 bodies larger than Sedna may exist in this faraway realm, which the astronomers dub the inner Oort Cloud.

How was the estimate, in bold above, calculated?

Unfortunately, the paper is not available on ArXiv (oh, what hardships we must overcome!), but I have found it here. In it, where the "900" figure is mentioned (2nd page, I believe), the authors (Trujillo and Sheppard) say that they ran simulations with the data already found and their additional findings, and found that 900$^{+800}_{-500}$ bodies larger than 1,000 km in diameter could exist. They also found that 430$^{+400}_{-240}$ relatively "bright" bodies could exist, as well. Notice that there is quite a lot of uncertainty there; they are rather cautious (and rightfully so) when announcing the results of their simulations.

Interestingly enough, they also mention a paper in that section, written partly by none other than Mike Brown (!) that also used simulation methods. The authors, Schwamb et al, concluded (in 5.4, as well as the conclusion) that 393$^{+1286}_{-264}$ and 74$^{+249}_{-47}$ objects greater than or equal to Sedna in brightness, not size, could exist (393 is for the "hot" population, while 74 is for the "cold" population). The reason Trujillo and Sheppard cited this paper is that their simulation results fit the results of the simulations by Schwamb et al., which is encouraging.

Summary: Trujillo and Sheppard arrived at 900 via simulations based on prior observations. There is a large amount of uncertainty in this number, but it does agree with prior simulations by Schwamb et al. While zibadawatimmy 's comment is perhaps the funniest I have seen, it is, fortunately, inaccurate. Those statisticians must be relieved.

I hope this helps.


Why aren't GPS clocks sufficient to synchronize VLBI observations? In very long baseline interferometry, it is typical to use a hydrogen maser atomic clock to sample data. However, given how expensive these atomic clocks are, it seems surprising that more observatories aren't using non-local atomic clocks, such as GPS clocks.

This Wikipedia article says "The antenna signal is sampled with an extremely precise and stable atomic clock (usually a hydrogen maser) that is additionally locked onto a GPS time standard." My question is, why not do away with the hydrogen maser entirely?

There is a presentation that covers this subject which is a regular at the International VLBI Service workshops. The most recent version is at High-accuracy Time and
Frequency in VLBI and the relevant derivation of needed accuracy is on slide 5. Basically for interferometery, you are trying to match up the phase of the waveforms (position along a sine wave in the simplistic sense) of the two signals from each of the dishes.

If we use an observing frequency of 10 GHz (10e9 Hz) and want to keep the phase coherent to ~10 degrees (out of the 360 degrees) after 1000 seconds integration then you need 10 / (360 * 10 * 1e9 * 1e3) or 10 / 3.6e16 or a clock stability of about 2.8e-15. Higher frequencies such as the 230 GHz mentioned in the comment will be even more demanding.

If you look at Allan Deviation/Variance plots, which measure the stability of clocks, such as the plot on slide 4 (and copied below):  this level of performance at 1000 seconds is only reached by hydrogen masers. The typical GPS Disciplined Oscillator will be at about 1e-12@1000seconds, see this recent comparison of 17 GPSDOs. With careful correction of the sawtooth error caused by the internal oscillator of the GPS (discussed at the end of the presentation) you can improve the typical 50 ns jitter of GPS to ~5 ns. Beyond this atmospheric/ionospheric effects come in and you need to use a dual-frequency GPS to take those out.


Speed of light through the ISM and Wavelength Articles written about fast radio bursts describe the signal's short-wavelength components arriving before its longer-wavelength components, suggesting energy-dependent time delays in the interstellar medium (ex. http://www.scientificamerican.com/article/a-brilliant-flash-then-nothing-new-fast-radio-bursts-mystify-astronomers/).

Why are lower-energy signals slowed slightly more than higher-energy signals when passing through clouds of electrons? Also, how do we know about this slowing effect if it appears to be so small (what objects or events have exhibited the time difference)?

My first-order logic guessed that higher energy radiation is slowed more, remembering the mnemonic "Blue bends best" from optics. I assume these intergalactic clouds behave quite differently from a solid prism.

You can find a neat description and some examples of the effect here. This is known as the pulsar dispersion measure. As you correctly say, waves with longer wavelength (lower energy photons) are delayed with respect to shorter wavelength radiation from the same phenomenon.

When electromagnetic waves travel through a plasma, they excite currents in the free charged particles. In such cases it can be shown (using Maxwell's equations) that the waves propagate with a relationship between their frequency $\omega$ and "wavenumber" $k = 2\pi/\lambda$ given by
$$\omega^2 = \omega_p^{2} + c^2k^2,$$
where $\omega_p$ is known as the "plasma" frequency and equals $(n_e e^2/\epsilon_0 m_e)^{1/2}$ for the electrons in the plasma (i.e. it depends on the electron number density $n_e$.).

Now, if you have a bunch of photons emitted as a pulse, the relevant velocity is the group velocity given by $v_g=d\omega/dk$. So
$$v_g = \frac{c^2 k}{(\omega_p^{2} +c^2 k^2)^{1/2}} = c\left(1 - \frac{\omega_p^2}{\omega^2}\right)^{1/2}$$

This converges to $c$ when the frequencies are high (wavelengths are short), but is slower when frequencies are low (wavelengths are long).

In terms of an intuitive physical picture, yes you could think that the refractive index is frequency dependent, but the difference here is that the reason for this dispersion is that waves travelling through a conductive medium are "lossy" - that is the currents induced also encounter resistivity and therefore the waves heat the plasma.


How is Earth's gravity modelled? The Earth does not have a defined geometric shape. Then how do space scientists who want to put a satellite in orbit model the gravitational field of Earth? Do they assume a single object with minute variations from sphericity, or as a combination of multiple objects of defined shapes? Or, are there different ways of doing it right?

Quick answer for now; the Earth's field has been more and more carefully measured by precise tracking of satellites over time. It stared with very careful visual tracking/timing and interferometrically tracking and recording the Doppler shift of the radio signals from late 1950's satellites like Sputnik and Vanguard and then many many others over the years, then on to pairs of satellites that track each others' distance GRACE and GRACE-FO in order to build an Earth Gravitational Model.

It's expressed as a bunch of coefficients of spherical harmonics. The monopole term is $GM$ the standard gravitational parameter, there is no dipole moment since the center is defined as the Earth's center of mass, and the quadrupole moment is expressed using $J_2$ and $J_{22}$, the first of which represents Earth's oblateness which is roughly a 0.1 % effect in low Earth orbit and is what makes Sun-synchronous and Molniya orbits possible.

The spherical harmonics of a given published model only work outside of (essentially) all of Earth's mass. Very near the Earth's surface (as opposed to in orbit) you would have to correct them for the shell of the mass of the atmosphere above you which is no longer affecting you as well as nearby mountain tops that rise above you.


What are some theories that explain the 11 year solar cycle? The Sun exhibits a periodic cycle for solar activity as manifested by solar activity (e.g. sunspots, solar flares). This activity is related to magnetic fields which emerge from the interior of the Sun. The magnetic fields are created by the solar dynamo. The Earth has a dynamo as well but the Earth's magnetic field is (mostly) constant in time. What is some current theories (assuming there isn't one already accepted) that explains for why the solar dynamo generates magnetic fields periodically?

It is linked to the solar dynamo and the dynamical behavior of the so-called $\alpha$-effect. One way to understand easily the solar dynamo and the the $\alpha$-effect with the hand is the following: you can generate a magnetic field by differential rotation (the $\Omega$-effect, $\Omega$ being the classical notation for the angular velocity). But this sole effect, twisting the magnetic field, would fold it so much that it will reconnect and dissipate. If you don't want to lose the magnetic field, you need a process to "regenerate" it, and a good way to do it is to create a poloidal component of the magnetic field. This poloidal component can be generated by convection in the solar interior; the magnetic field is, in the same time, twisted and pulled and that explains the solar dynamo.

Now, you can show that dynamo waves will propagate from the equator to the poles, and when the poles are reached, a reversal is observed. This propagation of dynamo waves is well observed by the sunspots patterns during the solar cycle.

Sources:

Parker (1955)
Proctor (2006); you can find it freely available here: http://damtp.cam.ac.uk/user/mrep/solcyc/paper.pdf


Are we sure there are no planets inside Mercury's orbit? Currently, most explorations about new planets are pointing to area outside Neptune's orbit, but how about inside Mercury orbit? Can we say for certain there are no planets inside Mercury's orbit?

Yes, Vulcanoids larger than 5.7 km diameter are considered to be ruled out:
A Search for Vulcanoids with the STEREO Heliospheric Imager.

An object sufficiently large to be called "planet Vulcan" would change Mercury's orbit in a detectable way by gravity.

So, there are two lines of evidence, that such a planet doesn't exist.

More to read in Wikipedia:
https://en.wikipedia.org/wiki/Vulcan_(hypothetical_planet)


Did the Moon ever approximate a geosynchronous orbit above the early Earth? Simulations suggest the early Moon formed about 20,000 to 30,000 km away from Earth which is less than the current geosynchronous orbit of 42,164 km.  The Moon is now 384,000 km away from Earth.  Does that mean the moon approximated a geosynchronous orbit for a time on its journey outward to its current orbit?  I recognize because of the Earth's tilt of 6.7 degrees the Moon does not orbit around the Earth's equator and therefore a geostationary orbit wasn't possible.

If the early Earth was a water world covered almost entirely with water at the time it formed, what would happen when the Moon approached geosynchronous orbit and applied its much more amplified gravitational force upon the Earth?  Would there be tidal bulges of water at both ends exposing land at the polar regions or would the water eventually make its way toward the Moon side possibly exposing or making a bulge of land at the opposite end of the Earth?

Maybe. The period of a 20,000 km orbit is about 7h 50m. Wikipedia says:

  [The Earth] would have experienced a day some five hours long after
  the impact, and the Earth's equator and the Moon's orbit would have
  become coplanar.

In the present era, the Moon's orbital plane is tilted by
5.145° to the ecliptic (the Earth's orbital plane); the obliquity of the ecliptic (relative to Earth's equatorial plane) is about 23.4°. However, as Wikipedia says:

  The Moon's orbit around Earth has many irregularities (perturbations),
  the study of which (lunar theory) has a long history.

The post-impact Earth was very hot, and had a magma ocean for a while. However, it soon had liquid water again because the high atmospheric pressure pushed the boiling point of water way up.

From Wikipedia:

  [From the Theia impact] a fair fraction of material should have been
  vaporized by this impact, creating a rock vapor atmosphere around the
  young planet. The rock vapor would have condensed within two thousand
  years, leaving behind hot volatiles which probably resulted in a
  heavy $\mathrm{CO_2}$ atmosphere with hydrogen and water vapor. Liquid water
  oceans existed despite the surface temperature of 230 °C (446 °F)
  because at an atmospheric pressure of above 27 atmospheres, caused by
  the heavy $\mathrm{CO_2}$ atmosphere, water is still liquid. As cooling
  continued, subduction and dissolving in ocean water removed most $\mathrm{CO_2}$
  from the atmosphere but levels oscillated wildly as new surface and
  mantle cycles appeared.

It is difficult to estimate the effects of the interaction between the early Moon and that primordial ocean.


How to improve efficiency in periodicity detection? Our data set has $10^4$ data points, but has a long baseline and many gaps.

If we bin the data, there would be $10^8$ data points ($[t,\rm {value}]$), but only about 1% are non-zero values.

How to improve the detection efficiency?

Is a multi-threading way possible (especially for Lomb-Scargle)?

For example, if I use LombScargle in astropy.stats,

freq, power = LombScargle.autopower(minimum_frequency=0.5, maximum_frequency=1.5, normalization='standard')

there are two problems:

It is slow and easy to get memoryError.
It can't utilize the computing power as much as possible.

So my question is about the efficiency. For the data set above, what is the best method?

AstroPy

Jake Vanderplas (Director of Open Software at the University of Washington eScience Institute) has written a Pythonic Preambulations post Fast Lomb-Scargle Periodograms in Python which by definition appears to be a good place to start to learn how to do Lomb-Scargle Periodograms... Fast... in Python.

I have not read it yet in full, but I've found other posts of his to be extremely useful.

He starts out by explaining the implementation in AstroPy which if you are going to be doing Astronomy, is probably a good thing to become familliar with.

SciPy

Alternatively, there is scipy.signal.lombscargle which you can try assuming you have a current version of SciPy installed but want to try something different than AstroPy or want to compare performance or results.

StackOverflow

I found over 40 posts containing 'Lomb-Scargle' in Stack Overflow which is a very good sign!

I notice that you've been active there and know how to ask well-received questions. Since your asking about speed and errors, StackOverflow is your go-to source! However, don't forget to be as clear as possible and quote all error messages and symptoms, and see if you can somehow add a short Minimum Complete and Verifiable Example or MCVE. That's hard when your problem results from a very large dataset I know.


Why the light curve goes down when the planet is behind the star? There is a video explaing the transiting exoplanet light curve — https://www.youtube.com/watch?v=RrusIZaWDW8

It is clear to my why the curve goes down when the planet is between the observer and the star, but I don't understand why the curve goes a little bit down when the the planet is behind the star. I was expected that this should not change the light curve at all.

Think of when the planet is at the "side" - there's a little bit of light from the planet (ie, reflecting off the planet) shining towards us.

Could it be due to that little bit of light - when the planet is behind the star, it no longer reflects towards us?

Maybe that's the effect you have in mind?


How much difference did the supply of "fresh" r-process elements make to the Solar System.? In one of the answers to this question @MartinKochanski made the interesting point that the abundance of elements heavier than iron (r-process elements) in the solar system is probably due to a fairly recent and close (at the time) neutron star merger -- a relatively rare event on a galactic scale.  This means the early solar system was enriched with "fresh" r-process elements, as compared to the "stale" r-process elements that would be typical. For instance the early solar system is conjectured to have been rich in uranium and thorium compared to a "typical" system.

So my question is what differences might this have made to the formation and evolution of the solar system? Would it have had any bearing on the origins of life on Earth?

The discussion in that answer is slightly misleading. Here's the current picture, taking into account the Nature paper by Bartos & Marka referenced there.

Elements heavier than iron up through rubidium (atomic number 37) are produced by the r-process in supernovae.
Elements heaver than rubidium are produced by a combination of the s-process in the asymptotic giant branch stage of intermediate-mass stars (i.e., masses of 2-10 solar masses) and the r-process in neutron-star mergers.[1]
The argument in the Nature paper is that when the solar system formed, a significant fraction of actinide metals with short half-lives (e.g., isotopes of curium and plutonium, with half lives $< 100$ million years) came from one recent, nearby neutron-star merger. (These are the "fresh" r-process elements.)
Stable heavy elements produced by neutron-star mergers would have come from the cumulative contributions of hundreds (possibly a few thousand) NS mergers over the history of the Milky Way, mixed over hundreds of millions or billions of years into the Galaxy's interstellar gas (these are the "stale" r-process elements).
Unstable heavy elements with long half-lives -- e.g., U-238 (4.5 billion years) and Th-232 (14 billion years) -- would be mostly in the "stale" category, and so would not be significantly affected by the presence of said recent, nearby NS merger. In fact, the Nature paper estimates that their hypothesized nearby NS merger would account for only 0.3% of the solar system's initial Th-232. Even U-235, with a half-life of 700 million years, would probably only be moderately affected by that one NS merger.

In general, then, the presence or absence of metals from a single nearby NS merger would have made very little difference to the formation of the solar system or to the evolution of life on Earth.

[1] See this figure for an approximate breakdown by element -- note that the "dying low-mass stars" label in the figure refers to the AGB stage of what I call intermediate-mass stars, which are "low mass" only in comparison to stars with masses $> 10$ times that of the Sun.


Any cheap (less than 100 euro) telescope for observing Jupiter? The moon is not challenging.

I would like to observe the big planets and their rings.

Jupiter is pretty easy. A little refractor with a 50 mm aperture (diameter of the primary lens) will show you the disk of Jupiter and the biggest two equatorial belts, and the four galilean satellites. Or Saturn and its rings - that's doable too. The image will be tiny, but you'll see those features - assuming the instrument is not junk and it can take a decent magnification.

However, that's all that you'll see at that level of investment. No small equatorial belts on Jupiter, or inter-belt swirls. With the Great Red Spot shrinking and becoming washed out each year, you may not even see that one. Also, Saturn will be disk + rings, that's it; no Cassini division, no ring shadow, or anything else.

If that's okay, then keep reading.

Availability of instruments depends on where you live. I'm more familiar with the market in North America, but EU can't be too different.

Surprisingly, a very decent cheap telescope is the Galileoscope.

http://galileoscope.org/

Yes, it's a DIY kit that you assemble yourself. But the primary lens is a surprisingly good achromat doublet; the design makes sense for an instrument like this; it accepts standard 1.25" eyepieces used for bigger telescopes; and the price is within your limits.

As an amateur optician and telescope maker, I was pleasantly surprised by this instrument. It works well up to about 100x, which is the theoretical limit for its aperture.

What will be difficult with this scope - and with any cheap refractor - is the mount. For astronomy you need a rock-solid mount, which is hard to do. The Galileoscope can be plugged into a standard photographic tripod - maybe you can find a good one second-hand. The site also gives suggestions for DIY mounts. Try different things and see what works. I've seen many commercial scopes, more expensive than this, that were basically unusable due to really poorly thought out (or poorly executed) mounts. Beware.

If the DIY route is not appealing, try some ready-made scopes. Here are a couple suggestions of vendors (I'm not affiliated with anyone):

http://www.telescope.com/Telescopes/Best-Telescopes-for-Kids/pc/1/459.uts

http://www.optcorp.com/telescopes/all-telescopes.html

Sort by price and keep digging. The mount will be a weak point, always, in this price range. Those refractors with tall, thin tripods tend to be very wobbly. It's a true hell to push the scope a bit this way, only to have the tripod rebound the other way - very annoying.

You may have better success with one of those short, stubby 76 mm table-top reflectors. Tweak the mount until it moves relatively easy, but not too easy. Hold it with both hands and guide it to track the planet. It may actually work better than one of those tall refractors. OTOH, the focal ratio for this reflector is only f/3.9, which is awfully short (it's hard to control aberrations for such a short instrument, unless you invest in high-end optics).

If you do get a refractor, get one of those with short legs. The nice thing about these cheap refractors is that they have a long focal length (large focal ratio), which helps keep aberrations low.

Frankly, there will be some limitations in this price range no matter what. It would help if you could buy from a store with a good return policy - just in case things don't work well for you.

Finally, you say "the Moon is not challenging"? It can be very challenging, once you get into exploring the tiny craters and so on. Just a suggestion. ;)


Are there clouds/clusters of gas in space that have red and blue color adjescent to one another? In artistic renditions of space scenes, there were stunning depictions of gasses or galaxies with bold red and blue colors immediately adjacent to one another, in contact as water is with oil in a container. I wonder, are there known arrangements of gas like that, with both ends of the color spectrum adjacent to one another? If so, what are some examples, what is known about them, and what causes the two colors?

One reason I'd call it into question is that I'd imagine the properties of boldly blue gasses are different than those of red gasses. Maybe that assumption is wrong because the property of color isn't necessarily associated with other properties. Another assumption would be that gasses with very different properties wouldn't find themselves collecting adjacent to one another in space. That assumption could also be wrong - I'm not an astronomer and only have basic understanding of physics and chemistry.

Images that you see of nebulae show more intense colour than you could ever see with your eyes. Most nebulae are too dim to for our colour vision to be very active when looking at them. Cameras are designed to try and mimic human sight, to produce an image with colours that is close to what we would see. Telescopes are doing science and the colours are there to enhance features that we couldn't otherwise see.

A telescope like Hubble will image a region at several wavelengths. Some of these wavelengths might be in the infrared or ultraviolet, beyond human sight. Then a colour is assigned to each wavelength, and the images are mixed. (See this description of how to achieve this with photoshop) This brings out details that would be lost in natural colour.

Now there are two main reasons that dust and gas in a nebula might shine. One is that it is reflecting the light of nearby stars, the other is that it has been energised and is emitting its own light.

Hydrogen gas will emit light with a red colour (in the right conditions), and some stars shine in blue-white light. If an emission nebula is next to a reflection nebula then you will get blue-on-red. A good example of this is the Trifid Nebula in Sagittarius. As in this photo by Scott Rosen

  The lower, reddish part of the nebula contains the emission nebula — hydrogen gas which is emitting red hydrogen light since it is being heated by the hot young stars within the nebula. The upper blue portion of the Trifid is a reflection nebula — it consists of dust particles that are reflecting the light from the hot stars contained within the emission nebula. Because the hot stars are mostly blue, we see the reflection nebula as having a blue hue.


Could matter be infinite? Perpetual motion and Big Bang theory seem to allude that the universe can’t last forever because matter and energy are dispensed.
Is it plausible the Universe has infinite matter?

Yes.  Currently it is at least plausible that the universe is spatially flat.  If that is the case then, barring global topological weirdness, it is also spatially infinite.  That would mean it contains an infinite amount of matter.
Disclaimer. I don't know how large the uncertainties on spatial flatness currently are: if Wikipedia is to be trusted $\Omega = 1.00 \pm 0.02$ ($1$ being flat) based on data from WMAP and Planck.


Symmetry in Hawking radiation? I have a question concerning how Hawking radiation works.

As I understand, a pair of particle-antiparticle may pop to existence near the event horizon (from a virtual particle).

Then it might happen that the negative particle (antimatter) gets captured by the black hole whereas the positive particle (matter) escapes.

This results in a net loss of the black hole mass. If you repeat this process an astonishing number of times the Black Hole might end up completely evaporating.

Am I right so far? (Even if the explanation is not very mathematical)

My question then is:

This assumes that when a pair particle-antiparticle is created, a bias towards antiparticles being captured and particles escaping must exist.

Otherwise, the mass of the black hole might fluctuate, but in average for big numbers of pairs, equal percentage of particles and antiparticles would be captured/escape.

So, is there really a bias needed for the black hole to loose mass, or is it something wrong about my understanding of Hawking radiation?

Thanks

This assumes that when a pair particle-antiparticle is created, a bias towards antiparticles being captured and particles escaping must exist.

No such assumption or bias exists.

As far as we know the universe treats anti-particles just like particles with different properties (e.g. charge is opposite compared to the corresponding particle).

  Otherwise, the mass of the black hole might fluctuate, but in average for big numbers of pairs, equal percentage of particles and antiparticles would be captured/escape.

You seem to be implying that anti-matter has negative mass.  All particles (including anti-matter) have either zero mass (like photons) or positive mass (like electrons).

So regardless of whether the anti-particle or particle escapes, the black hole looses mass (more precisely it looses energy).

  So, is there really a bias needed for the black hole to loose mass, or is it something wrong about my understanding of Hawking radiation?

Your understanding of the mass of anti-particles is likely the problem.  They have positive mass, just like particles, because they are particles.


Planets visible at night Stars give off light and are visible at nighttime but Planets do not give off light . So why are Planets visible at nighttime ?

Planets and moons reflect sunlight, as mentioned in this comment by AtmosphericPrisonEscape.

Now I'll add a bit more info to make this post more informative :)

That also means that just like the moon, the planets have phases, but they're just too subtle to see with the naked eye, and for any planet further from the Sun than the Earth is, we don't see all the phases.  They're full when the planet is furthest from us (on the other side of the Sun), so it'd be hard to tell the phase just from how bright the planet appears.

But through a telescope, you can easily see the phases of Venus, and you can tell it's smaller when it's full (because it's farther away):

Source: Wikimedia Commons


Could life survive on a planet orbiting a neutron star? Suppose an Earth-like planet orbits a neutron star, though specifically not a pulsar. Is there any possible orbit, around any possible neutron star, in which the planet could support human life?

It is possible that there could be planets where terrestrial-type life could survive orbiting a neutron star, but they are likely extremely rare and transient.

The habitable zone where liquid water can persist around a star has a range $\approx 0.7\sqrt{L_*}$ to $1.4\sqrt{L_*}$ AU, where $L_*$ is the luminosity of the star measured in solar luminosities. The exact limits are debatable and depends somewhat on the planet's properties, but this is close enough for this answer. A typical neutron star has a low luminosity, so the habitable zone will be very close and narrow.

Neutron stars cool down very rapidly at first. A young, million-Kelvin neutron star has luminosity around 0.18 solar luminosities, but a 10 million year old neutron star is down to 0.000026, and after a billion years it will be down to a millionth of the sun. This rapid cooling means that the life zone moves inwards from around 0.3-0.6 AU for the young star to 0.003-0.007 for the 10 million year star to 0.0007-0.0014 AU for the old star.

This means that a planet orbiting inside the zone at some time will become too cold before too long unless it is very close to the star since the zone drift slows down over time. It is hence unlikely that life would naturally be able to emerge on a planet like this, since planets that are in the habitable zone early only get a few million years before they freeze. The close planets that get a long period with decent temperature also have a period of extremely hot temperature before that, when most likely all volatiles are boiled away.

(Given that such a system has experienced a supernova in its past, volatiles will have been reduced even further).

That said, I don't see a fundamental reason a close planet might not be terraformed with volatiles added from elsewhere. If we consider a planet orbiting a $L_*=10^{-6}L_\odot$ neutron star $10^{-3}$ AU away (that is, 149,597 km from the star) it could maintain water. It would have a period of 16 minutes. There would be some relativistic corrections to the orbit, but they are minor (we are $10^5$ Schwarzschild-radii away).  It would be tidally locked (since tidal locking happens on a timescale $\propto a^{6}$), but currently there are a fair number of atmosphere models suggesting such worlds could remain stable/ Living on the shady side would also prevent the damaging UV and X-rays, which would likely strip away the atmosphere over time. The main problem is that there would not be any geodynamic dynamo to power a protective magnetic field; this is a planet that really needs it.

It is worth noting that there is a very different system around a neutron star that in principle (but likely not in practice) may be habitable, and that would be far away from an accretion disk. A quiescent accretion disk may have a luminosity about 1/3 of the sun, but when going active this increases by a factor of 1,000-10,000. So here one could imagine a planet orbiting at distance, getting powered by the accretion disk light. This will likely be too unstable to actually work over long periods and have the same hard radiation problem as the close one, but on paper (or in a suitable work of fiction) it is not impossible.


How to read 5 degree data of green line intensity When I open the 5 degree data of Green Coronal Emission line from this, I get a weird table which is not simple like the one for Coronal Index. I am only providing the table for 1939.

http://www.ngdc.noaa.gov/stp/space-weather/solar-data/solar-indices/solar_corona/5-degree-data/data/1939/1939

I can't provide the link for the formats used due to the restriction (max of 2 links until I have a rep above 10) but it's in the folder for 5 degree data.

I searched extensively connecting the formats (like I4, I2) to the datasheet but couldn't even scratch the surface of it.

I want the data for green coronal intensities and positional degrees in that respective year but can't find from this table. How do I find them?

The format of the files seems to be described in this file, the content of which I've included below.
The annual files (1939-1993) contain the green coronal emission line
530.3 nm observations.  The coronal intensities are given in millionths
of intensity of the solar disk (coronal units) and converted to the
photometrical scale of Lomnicky Stit Station at a height of 40" above
the solar limb.

Several stations were used in this database with Lomnicky Peak being the
primary station since 1965.

Reference:  Rybansky, et al, 1994, Solar Physics 152, p 153-159.

Data format:

 COLUMN  FORMAT   DESCRITPION
  1-  2    I2     Year, last two digits
  3-  4    I2     Month
  5-  6    I2     Day
  7-  8    2X     Blank
      9    I1     Coronal station code:
                  0 - Interpolated data
                  1 - Lomnicky Stit
                  2 - Sacramento Peak
                  3 - Norikura
                  4 - Kislovodsk
                  5 - Pic Du Midi
                  6 - Wenselstein
                  7 - Arosa
                  8 - Kanzelhohe
 10- 13    I4     Time of the observation in hours (or zero)
 14- 17    I4     Time of the observation in minutes (or zero)
 18-305  72(I4)   Coronal intensities (72 values) from 0 degrees to 355
                  degrees of position angle in increments of 5 degrees.

Using the first row data as an example...
DATE        SC  H   M   1   2   3   4   5   6   7   8   9   10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30  31  32  33  34  35  36  37  38  39  40  41  42  43  44  45  46  47  48  49  50  51  52  53  54  55  56  57  58  59  60  61  62  63  64  65  66  67  68  69  70  71  72
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
19390101    0   0   0   0   0   0   0   0   0   0   0   3   3   3   8   15  22  34  41  52  46  40  52  68  59  49  35  17  8   3   3   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   3   3   6   8   12  19  31  40  50  47  43  38  53  73  71  65  56  47  38  24  17  12  6   0   0   0   0   0   0

19390101 is the date (1939-01-01).
The first zero (Marked as SC) is the station code, the second the recorded hour, and the third the recorded minute (H/M). Most of this is 0 due to being interpolated data.
After this information, there are 72 remaining columns (numbered 1-72 above), which represents collected data consolidated in groups of 5 degrees, presumably starting at 0.
Given the amount of interpolation of old data the value may be questionable, but newer data appear to be more complete.
The data format might be a little overwhelming to work with, but you can organize it in excel quite easily - paste it in and each line break will form a row, then use the text-to-column function to break each data into its own column.


Connection between Earth, Moon, Theia, and asteroid belt? Are there other fragments aside from the Moon that resulted from the collision of Earth and "Theia" (as it's called in recent main-stream press stories)?  And if not, why not?  And what happened to "Theia" (like were did it go?)

Has there been any consideration or hypothesizing that Earth may have had a larger orbit (say between Mars/Jupiter) until a large body (Theia) bashed into it, sending Earth closer to the sun along with its new satellite, while also creating the asteroid-belt?  Could something like Ceres be the missing "Theia"?

I know it sounds a bit far-fetched, but then so does the whole concept of the Moon being born from a planet collision with Earth (with very little evidence to back it up, at least till recently).

Let's start answering your question in reverse. Ceres could not be "the missing Theia", because of its shape. If an object hit the Earth at an angle (as is currently thought), it would be pretty deformed, if it managed to stay together. If it hit the Earth head on . . . Well, it would almost certainly not survive. So if it did survive the impact, it would be in pretty bad shape. Where would such an object go?

Let's investigate the idea that it went into the asteroid belt. That begs a question: Where is it today? Ceres is the largest object in the belt (and we've already ruled it out). Theia had a mass about that of Mars (see Canup (2012)). It's safe to say that if Theia survived with much of its mass intact and then disappeared into the asteroid belt, we should have observed it by now.

Now for your belt-formation theory. It is, admittedly a very cool idea. The mass of the asteroid belt is about 4% that of the Moon - so it could be the result of the planetary crash. And some theories say that the belt is, in fact, the remains of a planet. There is, of course, one problem - the different composition of the asteroids. There are three types, differentiated by their composition - C, S, and M. So it is very unlikely that they came from the same planet. Meier et al. (2014) show evidence that indicates Theia's composition would have been more like Earth, containing enstatite chondrites instead of carbonaceous chondrites.

Finally, let's look at your idea about the Earth having an orbit farther out. This one seems to have a good chance of being right. The early protoplanets in the Solar System were very - very - unstable in their orbits because of the frequency of collisions. So the Earth could have been in a different orbit early on, and then migrated inwards or outwards, though planet-planet scattering or perhaps gas disk migration.


How should exoplanet discoveries affect SETI? Given that SETI was created before the discovery of exoplanets, has this changed how they search? Would it make more sense to focus efforts on discovered earth like exoplanets?

Well you are correct and of course that kind of research has been going on. The SETI programme targeted a set of Kepler stars with evidence for a planet (or planet candidate) in the "habitable zone". The initial searches for narrow-band radio emission (at 1.1-1.9 Hz) appears to have concentrated on those systems containing multiple planets.

The (null) results for $\sim 100$ system were published by Siemion et al. (2013). The observations were not sensitive enough to detect the sort of "radio chatter" emitted from our planet, but might have detected some sort of deliberate signalling beacon.

A slightly more imaginative approach, which I'm struggling to find the details of, is that one can observe Kepler multiple planet systems when one calculates that two planets are in conjunction as seen from the Earth. There is then the chance of seeing communications from one planet to the other. (e.g. http://adsabs.harvard.edu/abs/2014ebi..conf..5.2K ).

In a more general sense, the exoplanetery discoveries have provided great impetus to the SETI programme. It is now clear that planetary systems are common and present around most stars. It is also now apparent that small planets are common. New possibilities are also being considered. Most exoplanets in habitable zones are orbiting stars smaller and less luminous than the Sun. No exact Earth analogues have yet been found. This has led to an explosion of activity investigating planetary atmospheres and habitability around K- and M-dwarf stars.


Fraction of circumpolar stars to all stars visible in a location We know that circumpolar stars are stars that don't rise or set. If we are in a location with $\phi = 35^o N$. What is the fraction of circumpolar stars to all stars visible in this location?

I know the declination of this stars is $\delta > 90 - 35 = 55$ But I don't know what is the degree of all stars visible in this location and is the fraction as simple as dividing this two angles or not?

The answer depends on how detailed you want to be. For example,

Do you want (a) an approximate answer by assuming the stars are uniformly distributed, or (b) to use the real distribution of stars?
Do you want to know the answer based on (a) the number of stars visible throughout the entire year, or (b) just a given night, or (c) just an instant of time?
The magnitude limit of your sky affects how many stars are visible. More stars are visible from a dark site in the country than in a city.
The other factor that has a small effect on the answer is whether you want to include atmospheric extinction or not. Stars near the horizon appear to be fainter than their published magnitude due to the atmosphere. Some stars close to the southern horizon (in your example) are not visible because they do not rise high enough.

Combination 1(a) and 2(c) [uniform distribution, an instant of time] can be calculated based on the area of sky that is circumpolar versus the area for half of the sky. (As @zephyr indicated, you need to use steradians to calculate the area of the sky.)

Combination 1(a) and 2(a) [uniform distribution, entire year] is also easy to calculate from the area of the sky that is circumpolar, and the area of the entire sky minus the area of the sky that never rises (declination less than -55 for your example).

Any other combination requires using a catalog of star positions to "count" how many stars are visible at a given instant of time. In other words, based on the day and time (or the local sidereal time),

calculate the altitude of each star.
if above the horizon, calculate if it is bright enough to be visible given the sky conditions and (optionally) atmospheric extinction
if visible, tally the star as circumpolar or not
after counting all of the stars, perform the final calculation of the ratio.

Such a calculation would be easy to perform in a spreadsheet using a catalog such as the Yale Bright Star Catalog (which may not be complete down to the faintest stars visible). I did a similar calculation many, many years ago. See the May 1994 Sky & Telescope magazine, pages 86-88.


Mass distribution in the early universe The latest big quasar find at ~12.8 bn LJ with an estimated mass of 12 bn M☉ (see e.g. http://www.newscientist.com/article/mg22530104.000-ancient-black-hole-had-an-inexplicable-growth-spurt.html) puts current black-hole-forming hypotheses into question. As the linked article explains, scientist doubt that most of the matter could pass into the hole in such a short time due to the outward pushing force of the radiation created by the material falling into it.

My question is rather, was there enough matter in such a close vicinity of a BH that it could grow to this size? At 900 million years from the big bang I have the feeling that there is hardly enough time for the required matter to even pass through the volume around the BH where accretion can happen.

It is unfortunate that the usual poor journalism labels the growth of the black hole as "inexplicable" and then further down in the article refers to some possible explanations.

The basic problem is a growth timescale one. Radiation pressure introduces a negative feedback, such that there is a "theoretical" maximum for spherical accretion called the Eddington limit, which occurs when the quasar is radiating at its Eddington luminosity. The shortest growth timescale is achieved if the efficiency with which mass is converted to luminosity is low; but if it's too low we wouldn't see the quasar at all. This is the crux of the problem. You can look at this Physics SE answer for some more of the details.

The thing is there are ways and means by which this limit can be exceeded - non spherical accretion for one - so there are lots of ideas about how this can be achieved. Another possibility is that you start off with a seed black hole that is pretty big to begin with, perhaps as a result of a merger. Or the quasar could have been less efficient in the past and is more efficient now, which is why we can see it.

Is there enough matter? Well, yes, galaxies have masses that can be much bigger than the mass of this black hole. They are rare, but of course so are > 1 billion solar-mass black holes, and these tend to be the only ones that we can see at distances of >10 billion light years.

One way of assessing the feasibility would be just to ask what a freefall timescale would be. If you have say $10^{11}\ M_{\odot}$ in a sphere of radius 10 kpc (I am just using typical sorts of numbers for a big galaxy), then the average density is $5\times 10^{-22} kg/m^{3}$ and has a freefall time $\sim (G\rho)^{-1/2}$ of 200 million years. Of course there are other problems, like shedding angular momentum, but it looks like this timescale is short enough for gravity to do its thing (in the absence of radiation pressure).

Of course the short answer to your question is that yes, there must be enough time, because this is just the latest in a population of such objects. We know that quasars with supermassive black holes have formed within a billion years after the big bang.


Affordable night sky photography As an amateur with limited budget, I'd be interested in taking photos of the night sky, trying to capture more detail than human eye armed with a lens of comparable parameters to what I have in my camera normally could see. I doubt I'd ever get down to details as fine as Jupiter's moons, but I'd hope to see detail of some nebulae I have a hard time seeing through my inexpensive telescope, stars too dim to notice in less-than-perfect conditions etc. I'm interested in taking full-sky images just as well as zooms on specific objects too.

Currently, I have a lower-end SLC camera, with two lenses - good sharpness though lower aperture with 50-120mm focal length, and a wide-angle, high-brightness one (about 12-50mm) currently. Firmware hacks allow me to take photos of arbitrarily long time, and I have the remote to start and stop it without touching the camera, and generally software-wise the camera is quite powerful. One of the lenses (the longer focal length) is of "standard professional" quality level too.

Is this sufficient to get started? If so, what kind of settings should I use? If not, what other kind of entry-level equipment would I need to obtain/build on budget to get started with night sky photography?

For long exposure pictures you need to have a motorized mount for your camera. The earth's rotation will lead to streaks otherwise. An affordable way to do this is to use a standard tripod with a star tracker on top. There is a variety of products like:

iOptron Sky Tracker
Vixen Polarie

They cost around 400-500 USD and are fairly small. You need to align them to the sky's north pole (using the star Polaris in Ursa Minor), and then they will rotate your camera as the earth rotates during the night.

Also, you may want to start with your 50mm lens first. The greater the focal length, the higher the magnification, and also the more visible the tracking errors and misalignments of the sky tracker and camera system will be. Be prepared to spend quite some time trying to get good results!


Are all impact craters circular? I'm aware for over 100 years the reason has been known how impact craters are almost always circular, but my question really is has there been any recent insights in the area? What other topologies are physically possible?

There are elongated craters from grazing impacts, and these can sometimes get rather odd shapes like Schiller on the Moon and Orcus Patera on Mars (the later is a bit contested, it might have formed from something other than an impact).

Double impact craters is another possibility. Some craters occur simultaneously (more direct picture). Splitting impactors is likely the one way of getting truly complex topologies. Most of the time they tend to follow each other, producing crater chains.

If terrain shifts craters may be deformed, and depending on details in the rock they may also get odd shapes. Meteor Crater in Arizona is somewhat squarish.

Craters cannot generate that complex patterns since they are just explosions out from a center and impactors are usually single objects or random clouds. In principle one could arrange asteroids along a trajectory to spell out letters upon impact, but it is an exceedingly unlikely configuration so most impacts will only manage 'O'.


If we had the right technology could we see a distant star in detail? If, for instance, Hubble's camera was a $10^{300}$ times better, could it see a distant star in details, or is there a limit to the amount of light that reaches the Earth's orbit or perhaps does the light get distorted on its way so you must come closer?
Is there a limit to the number of light beams that hit a telescope camera's "eye" or if you built a huge telescope would you then be able to see that distant star in high detail (for instance, see an earth size satellite around Betelgeuse's orbit).

The only limitations would be related to building an instrument large enough.

There's a limit to the size of the finest detail a telescope can see. "Size" here is angular size, the angle that the detail is covering in the visual field. For a given wavelength of light, the smallest angular size depends on the aperture (diameter) of the telescope in a linear fashion: double the aperture, and you cut the angular size in half.

Rule of thumb: 100 mm aperture has a resolving power of 1 arcsecond, for visible light (not exactly, close enough). Therefore 200 mm aperture has 0.5 arcsec resolving power, etc.

By building larger and larger telescopes, the angular size of the smallest details becomes smaller and smaller. But are there any limits here?

There aren't any cosmologic mechanisms that I'm aware of that would prevent telescopes from operating at arbitrarily large resolving powers (arbitrarily small angular sizes). But we can't build arbitrarily large instruments.

The biggest monolithic (single mirror) telescope has an aperture of 8.4 meters. Using segmented mirrors, the aperture aim of the current largest project is 30 meters - the Thirty Meter Telescope, still in a very early stage, completion date unknown.

Beyond that size, large scale interferometers are the only option, with synthetic aperture varying from dozens of meters to kilometers, and larger projects are being discussed.

In theory, very large scale interferometers could be built in space. But there might be inherent limits for the aperture of such systems. Anyway, this is hypothetical.

EDIT: Giant telescopes in general do not require a continuous surface of glass. Interferometers almost by definition do not use a single reflector. You could build a gigantic telescope or interferometer where the active reflective surfaces are just these tiny chunks at the periphery of a huge perimeter. Most of the area in between is empty / unused.

EDIT2: The performance of the "tiny chunks" reflector is not the same as the performance of the full size mirror, as mentioned in the comments.


How to mathematically describe a planet's eccentric orbit? Is there a formula for describing a planet's position along an eccentric orbit over time? I am particularly interested in the elliptical case (0 < e < 1).

I understand that the planet has a higher velocity closer to the perihelion. Ideally I would like to be able to calculate a planet's position on it's orbital path in relation to how far through it's orbital period it is, if that makes sense.

Alternatively, is there software (that doesn't require a university's budget) to work this out?

The formula is Kepler's equation, but to understand it you need to know three values:

$M$ is the "Mean Anomaly". It increases linearly from 0 to 360 over the period of one orbit, measured from periapse to periapse. So if a planet has a period of 100 days, then the mean anomaly at day 0 is 0, at day 50 it 180degrees, at day 25 it is 90 degrees. This is only the position of the body that is in a circular orbit. It is the first stage of calculating the position of the body in an elliptical orbit.

$E$ is the "Eccentric Anomaly". This is also not the actual location of the planet. It is defined by $\cos\,E = x/a$ where $x$ is the x-coordinate of the of the planet, measured from the centre of the ellipse not the focus, and $a$ is the semi-major axis of the ellipse

$T$ is the "True anomaly", the angle between the actual position of the planet and the line through the major axis of the ellipse, measure at the focus of the ellipse (ie at the sun).

These are related by the Kepler equation. $$M = E- e\sin E,$$where $e$ is the eccentricity of the ellipse. You can calculate M directly, then solve this equation (numerically) to find E. Providing the eccentricty isn't very close to one, Newton's method is effective at finding a solution to this equation.

Then the true anomaly is given as
$$
T   = 2 \, \mathop{\mathrm{arg}}\left(\sqrt{1-e} \, \cos\frac{E}{2} , \sqrt{1+e}\sin\frac{E}{2}\right)
$$
(where $\mathrm{arg}(x,y)$ is available as atan2(y,x) in many programming languages)

It is possible to use the Eccentric anomaly to calculate the coordinates of the planet directly using $\cos\,E=x/a$and $\sin\,E=y/b$ where a and b are the semi-major and -minor axes of the ellipse.

The word "anomaly" doesn't mean "false result", but "angular deviation (from periapse)" and has been in use since Kepler's time.


How could we tell if the Universe is infinite? I've heard from a number of people that the Universe is infinite, and from my perspective, I can't see how this can actually be known, especially given that the Universe started out a finite size and grows at a finite rate.

  Regarding the possible similarity of my question: I want to know how
  it could be and how we would know, rather than if it is. This is
  significantly different.

We don't know for sure, but it certainly fits into our theories. There is, of course, no way to actually test if the Universe is infinite, but right now we think it is likely.

Also, if you read my updated answer on your other post, the Universe has always been infinite in size. I explain over there how it actually works: space is created in between everything, and thus one could say the Universe is expanding.

These objects can actually drift away from each other faster than the speed of light. That is, light from them eventually won't make it to us, since they'll be drifting away too quickly.

Now, this doesn't actually go against Einstein's theory that the speed of light is the fastest thing in the Universe. Einstein said that nothing can travel through space faster than light — but here, space itself is actually being created between the objects. Distances are increasing because space itself is dilating, and thus we can drift apart from other objects faster than light.

Really, there is no limit (as far as I know) to how fast we can drift away. Farther objects will keep drifting faster and faster away, since our gravity has a much weaker effect on them.


How much water on the earth is made from hydrogen comes with solar wind? A great fraction of solar wind is protons or Hydrogen ions. Some part of it is captured by Earth. I am wondering how much water in my half a liter cup is made of hydrogen from the Sun.

Here is a way of getting a rough upper bound: the density $\rho$ of the solar wind is about 4 atoms per cm$^3$ (that we can assume are hydrogen), and it travels at $v=400-500$ km/s. That means that were the Earth to catch every atom directed at it it would gain $\pi R_\oplus^2 \rho v$ atoms per second, $2.0$ to $2.5\cdot10^{14}$. Which gives us a mass flow of $3.4$ to $4.3\cdot 10^{-13}$ kg/s. Earth has been around for 4.5 billion years, so over that time at most about 49 to 61 tons of hydrogen have arrived.
The total amount of liquid water on Earth is $V=1.386\cdot 10^9$ km$^3$ (which doesn't count the sizeable amount of water bound in rocks). About 11.19% of this mass is hydrogen, so the weight of hydrogen is $0.1119 V\rho_{water}=1.55\cdot 10^{20}$ kg. So the ratio between hydrogen from Earth and from the sun is about $2$ or $3\cdot 10^{15}$.
Still, in a half liter (about 27.7 mol water) there are $3.33\cdot 10^{25}$ hydrogen atoms, so there should be about 11 to 16 billion solar hydrogen atoms in the cup if this estimate is right.
Maybe Earth's magnetic field captures more atoms, which might increase the value. Likely most solar wind is instead pushed away and the true value is much lower. But I am pretty confident there are some solar atoms in the cup.


Is there a preference for orbital planes of planetary systems to be aligned with the galactic disk? I'm not sure for how many planetary systems we have a fairly good idea of the inclination and "argument of periapsis" of the orbital plane. I'm assuming there are a number of them that we can estimate due to co-planar assumptions in a multi-planet system or some other means.

I am asking if there are enough "known" orbital planes that we can have some statistical idea whether or not the orbital planes are somewhat aligned with the galactic plane or whether local forces would dominate galactic orbit angular momentum. (I'm thinking the latter is true.)

We would have to keep in mind of course that a system "directly above" us and parallel to the galactic plane would be "face on".

Unless I'm missing something we know very little about the orbital planes of individual systems.

Most exoplanetary systems (all of the transiting ones, whereas it is just a strong bias for the doppler-discovered planets) are discovered because they have high orbital inclination. This means that the planet's orbit takes it across the stellar disc as we view it. However, the entire arrangement can be rotated around the line of sight by any angle and this would give exactly the same observational signature.

That being the case, the only test I can immediately suggest is that if the planes of planetary systems were preferentially aligned with the galactic plane (ours isn't), then it would result in a lower planetary detection rate for stars in directions perpendicular to the Galactic plane in comparison with stars lying in the Galactic plane.

There may be some mileage in pursuing this, though I don't know of any results. For example the main Kepler field was directed just out of the Galactic plane, but the K2 fields are at a variety of Galactic latitudes along the ecliptic plane. If one could normalise somehow for the types of stars observed in each field and the differing data qualities then it might be possible to look at the "planetary occurrence rate" as a function of Galactic latitude.

The "all-sky" ground-based surveys are less suitable for this because they tend to avoid the Galactic plane because it is too crowded with targets.

The question can be more broadly addressed by looking at visual binary systems - i.e. those where you can resolve the components and study the orbit. There, it is possible to say something about the orbital planes of individual systems. A paper by Agati et al. (2015) has looked at this. They collected 51 systems with the requisite information, that are close to the Sun. They found only weak evidence for any anisotropy of the orbital poles and this weak alignment was closer to the ecliptic pole than the Galactic pole.