Question about doppler effect We know that we can use Doppler effect to determine equatorial velocity of a planet. We send a electromagnetic wave with wavelength $\lambda_0$ to an equatorial edge of planet and we get wavelength $\lambda$. So we can use Doppler Effect formula, (assuming $v \ll c$)

$$\frac{\Delta\lambda}{\lambda_0} = \frac{v_r}{c}$$

such that $v_r$ is radial velocity. But $v_r$ = $2*v_p$. And $v_p$ is equatorial velocity of planet.

I want to know where does the $``2"$ comes from. Someone said it is because of receiving the signal and then reflecting it. I want to know the math behind this not just to say receiving and reflecting. Why do we need to multiply the velocity by 2 to get the radial velocity?

Before I start, let me just say that this topic is vastly more complicated than you've presented and what I will be showing. The trouble here is that ultimately, everything you've done and I will do (to a lesser extent) uses approximations and assumptions. Because of this, it can sometimes be hard to understand the true underlying physics when looking at only the "simple" versions of these equations.

You say that the you've heard that the source of the two is because of receiving the signal and reflecting it, which is entirely true. Let's consider the simpler case of a police officer using radar to track a car's speed (just so we don't have to drag in the more complicated concepts of orbital motions of satellites, dispersion relations, etc.) You have a police officer standing by the side of the road with a radar gun, shooting radar waves of some frequency $f_o$ at the car as it recedes from him (note I'm going to use frequency but it is completely analogous to wavelength, assuming a simple dispersion relation). That wave travels to the car and the first peak of the wave hits the car. The car then has time to move slightly forward by the time the next peak hits the car. From the car's perspective, because of his relative motion to the police officer, the frequency of the wave hitting the car has been Doppler shifted to a new frequency, $f_1$. This new frequency is given by the Doppler equation

$$\frac{f_1}{f_o} = \left(1+\frac{v}{c}\right)$$

Note that this equation is the same as what you presented, except that it uses frequency instead of wavlength. So now, our car has been hit by these frequency shifted waves. It will now reflect those waves back at the police officer. The process of reflection effectively makes the car an emitter of the frequency $f_1$ but since it's obviously moving with respect to the police office, the police officer receives a Doppler shifted frequency, which I'll call $f_2$. Since the speed between the car and the police officer is still $v$, we know that the Doppler equation for the frequency shift of the reflection is given by

$$\frac{f_2}{f_1} = \left(1+\frac{v}{c}\right)$$

Now the police officer receives back a frequency $f_2$, but he originally emitted $f_o$, so he wants to know the velocity which caused the shift from $f_o$ to $f_2$. Fortunately, this can be determined by combining the two shifts above.

$$\frac{f_2}{f_o} = \frac{f_2}{f_1}\frac{f_1}{f_o} = \left(1+\frac{v}{c}\right)\left(1+\frac{v}{c}\right)$$

$$\frac{f_2}{f_o} = \left(1+\frac{v}{c}\right)^2 = 1 + 2\frac{v}{c} + \frac{v^2}{c^2}$$

At this stage, we make the assumption that $v^2/c^2 \ll v/c$ and this second order term can be safely ignored. We then wind up with

$$\frac{f_2-f_o}{f_o} = \frac{\Delta f}{f_o} = \frac{2v}{c}$$

What the police officer measures by using the radar gun and applying the Doppler equation is $2v$, but of course we know this to be twice the speed of the car, so of course to get the actual speed of the car, the speed measured by the radar gun must be divided by two. As you can hopefully see, this 2 comes from the double receive and reflect of the radio waves by the car. The same exact principle holds whether we're talking about radar guns and cars or satellites and planets.

Note: As I said at the start, thing's are vastly more complicated than I've presented. For example, we assumed that $v^2/c^2 \ll v/c$. In reality, to get a more correct answer, you wouldn't want to drop this term and you'd find that $v_r \ne 2v_p$ as you've indicated in your answer, and there's actually more to it than that. This is all further exacerbated by the fact that the doppler equation I started with already had some assumptions built in. The fully correct equation is more complicated than that.


Why different specifications for telescopes and binoculars? Sorry for the noob question but I don't seem to be able to find the answer on the internet.

I've been looking through some telescopes and binoculars and noticed that shops typically give different specifications for the two groups. For example, for telescopes I'll often see focal length, aperture ratio, or limit value, but I haven't seen any of those for binoculars; for binoculars, on the other hand, they'll mention exit pupil, field of view, or glass material, none of which I've seen given for a telescope. Why is that? Are the two devices so different as to be incomparable?

With a binocular, all its optical components are fixed -  the user can't change them.  What's important for the user to know is the size of the front lens, which determines the brightness (and in theory sharpness) of the image, the magnification, and the field of view.  These are all useful things to know.

A telescope has an interchangable component, namely the eyepiece.  The choice of eyepiece determines:

The magnification (= focal length of telescope divided by focal length of eyepiece)
The field of view, which depends on magification and design of the eyepiece
The exit pupil size (diameter of objective divided by magnification)
and a few other things

Hence, based on a particular eyepiece type you can work out the same information as you get with binoculars.


Strange object crossing the sky Was stargazing last night and spotted an object traversing the sky from roughly East to West. It had a bright light that was intermittently flashing but with three different intensities. The "third" flash was really bright like a camera flash. Never seen a plane with lights this bright and was not a satellite (like any that I've ever seen), nor was it the ISS as that was not in this part of the world at this time. Any ideas anyone?

It seems you saw a Tumbling satellite. They don't have a constant brightness. This is a result of satellite's rotation. Learn more: http://www.satobs.org/tumble/tumbleintro.html


Why are there uneven bright areas in this photo of black hole? In the recently released photo of a black hole shown above, which was created by using data from EHT, why is the lower region brighter than the one above? Is it because of the rotation of the accretion disk? Also what is the orientation of the accretion disk? Are we looking at it head on?

No, you aren't seeing the shape of the accretion disk. Although its plane is almost that of the picture it is far larger and fainter than the ring that is seen. The reason for this asymmetry is almost entirely due to Doppler beaming and boosting of radiation arising in matter travelling at relativistic speeds very close to the black hole. This in turn is almost entirely controlled by the orientation of the black hole spin. The black hole sweeps up material and magnetic fields almost irrespective of the orientation of any accretion disk.

The pictures below from the fifth event horizon telescope paper makes things clear.

The black arrow indicates the direction of black hole spin. The blue arrow indicates the initial rotation of the accretion flow. The jet of M87 is more or less East-West (projected onto the page), but the right hand side is pointing towards the Earth. It is assumed that the spin vector of the black hole is aligned (or anti-aligned) with this.

The two left hand plots show agreement with the observations. What they have in common is that the black hole spin vector is mostly into the page (anti-aligned with the jet). Gas is forced to rotate in the same way and results in projected relativistic motion towards us south of the black hole and away from us north of the black hole. Doppler boosting and beaming does the rest.

As the paper says:

  the location of the peak flux in the ring is controlled by the black hole spin: it always lies roughly 90 degrees counterclockwise from the projection of the spin vector on the sky.


Observing Comet 45P - reliable sources of position data Yesterday night I have tried find comet 45P, that should be still easily visible with my equipment at hand, however I was unable to find it.
I used data provided by this website.

Today I've checked the website's data against the Minor Planets Center's data loaded into Stellarium (the program) and found the predicted positions of the comet to differ by more than 7°. That is totally unacceptable when trying to find the comet with a FOV of a few tens of arcminutes (the finder presumably is too light-weak).

So does anybody know a definite, reliable source for the position data of 45P? I'm gonna try again today anyway with the positions provided by Stellarium, but it might be good to have another alternative.

The best way to get ephemeris of solar system bodies is to use JPL HORIZONS:

http://ssd.jpl.nasa.gov/horizons.cgi

You put "Target Body" as 45P, your "Location", "Time", and click "Generate Ephemeris". This will give the RA/DEC information for time by default.

If you want some advanced data other than the listed, e.g., phase angle, V-magnitude, RA/DEC change rate, etc, you can tune "Table Settings". I usually skip daytime and airmass < 2.

There can be some mismatch between the ephemeris and real observation. This is not because JPL HORIZONS gives wrong calculation, but because the reported observation data is not perfect. It calculates the orbit of the target from previously reported observational data to IAU MPC. IAU MPC provides the orbital elements inferred from that reported data, and these elements are used to generate ephemeris from JPL HORIZONS. So if the report is slightly imperfect and thus the inferred orbital element differs from the true value by a slight amount, it may cause the mismatch at a time far from the original observation. This is the reason we need more and more data continuously and update the catalog.


Is our Solar System moving independently through our galaxy? I would like know if our Solar System is moving independently through our galaxy or is stuck on the Orion Arm which is revolving around our galaxy? Is our solar system an onboard passenger of the Orion Arm?

The arms of a spiral galaxy are "gravity waves"  Stars orbiting the galaxy are pulled towards them and speed up as they approach, and then are slowed down as they leave. So an arm of the galaxy is not composed of the same stars.  The arm is, in some ways, like a fountain, because it can maintain its shape even though the stars in it are constantly changing.
So the sun will move in its orbit out of the arm and into another, just the same as other stars.


How much light comes from quasars in an AGN? From what I've been able to gather, the very bright spot at the center of galaxies is produced by both the central quasar and a dense cluster of millions of stars that surround it. How much of the central light is produced by the quasar, and how much comes from the star cluster?

Active Galactic Nuclei (AGNs) are very luminous galactic nuclei. In Quasars, a type of an AGN, the nucleus outshines the host galaxy. Therefore the main source of radiation is the nucleus, not the stars and the host.


How many stars are there in a Globular Cluster of 10^5 solar masses? I was wondering whether there is an easy way to approximate the number of stars in a Globular Cluster (GC) with 10^5 solar masses.

Can one, for instance just assume the GC is made of sun-like stars and therefore has 10^5 stars?
Or is this too simple?

Thank you for your answers!

The mean mass of the initial mass function is ~0.3 solar masses (Maschberger 2013).  A GC with 10^5 solar masses will therefore have ~3 x 10^5 stars in it.  Since the most massive stars will already have died in a GC, the mean mass of the present-day mass distribution of the GC will actually be slightly lower, implying a slightly larger number of stars, but this will be a relatively small correction (certainly not more than a factor of 3).


What is the difference between a neutron star and a white dwarf? What is the difference between a neutron star and a white dwarf? I know that both are very dense even if they go through different phases.

In a neutron star, the force of gravity is strong enough to press the protons and electrons together to form neutrons {1}, White dwarfs are only very compact. With even more mass, you get a black hole.
All three types are outcomes of star death, when the failing fusion in the middle of a star is no longer able to counteract gravity. What a star become when it collapses is depending on its mass.


Have any planets been discovered in Orion constellation? Orion constellation is a very famous constellation. It can be seen throughout the earth. Orion looks like a warrior with a weapon and skirt. I see this constellation daily and wonder, if there is a planet there.

Yes, there are  exo-planets discovered in the Orion constellation, for instance, the stars HD 37605 and HD 38529 both have two known planets.

HD 38858 is also interesting, as it has a known planet, and also a disk of comets.


Shouldn't all moon orbits be inherently unstable? A hypothetical simulation starts with a planet called Earth in a perfectly circular orbit around a star called Sol. There is also a moon called Luna in a perfectly circular orbit around Earth.

Question:
Can Luna's orbit remain perfectly circular or does it become elliptical over time? At what point does it become stable? Never?

Thinking:
The distance between Sol and Luna is not constant, it fluctuates as Luna orbits Earth. Therefore the pull sunward also fluctuates. Shouldn't this cause all moon orbits to eventually deteriorate into longer and longer elliptical orbits until they collide with the planet they orbit?

The answer to your simple question is that while a circular orbit could remain in the absence of any other gravitational fields, in reality you'll have an ellipse and it can be stable from the beginning, remember stable has a wide range of values.

If you allow drag through tidal forces etc then the orbit will change, eg our Moon is slowly getting further away from us, but we still consider this stable over realistic timeframes.

Additionally, the actual orbit will track out a sort of helix as the planet travels round the sun.


During what time of the year can Centaurus be seen from Tokyo? During exactly what dates of the year can the Centaurus constellation be seen from Tokyo, Japan? And what is the ideal date for viewing it from Tokyo?  I found on a website that it can been seen May-June in New Zealand, best visible at 21:00 (9pm) during May and that the midline is May 20th.

(According to the Japanese Wikipedia page for Centaurus, only from Okinawa can the entire constellation be seen; in Tokyo, apparently only part of it can be seen.)

You can use an application that shows the stars at any time and location. Stellarium is free, has nice features and isn't too complicated. You can select a star/constellation and flip through the time of day and through the days to see how its elevation changes over time.

In general, everyone with the same latitude sees the same sky at a certain date (assuming an equal elevation, which counts if you're standing on Mt. Fuji). So those dates for New Sealand are irrelevant for you.

Using the StarmapPro application, I made the following chart for your location (Tokyo) showing the elevation of Centaur at midnight for each month.

As you can see, it's visible from halfway march until the first half of may, with its maximum around the first week of april. It comes barely above the horizon, so you'll need a clear view to the horizon and try to avoid light pollution (good luck with that in Tokyo ;-)


Could ancient tectonic activity on Mars be due to it having oceans of sulphuric acid? Could Mars' ancient tectonic activity have ceased due to its oceans having been composed of sulphuric acid, which warmed up and evaporated due to the Theia event as ice chunks landed on its surface? Strongly conductive liquid sulphuric acid on early Mars' surface could have been responsible for generating its core spin causing tectonic action which then all died off as a result of its oceans evaporating and thus removing the conductivity responsible for the core spin, whereas Earth's oceans didn't evaporate (because they are water) so can phase to ice ages to prevent overheating.
Why else would Mars have evidence of ancient alternately polarised extrusion from now ceased tectonic activity?

First, tectonic activity have not ceased on Mars, although the planet is in a "stagnant lid" stage, which represents the last stage before a body becomes tectonically dead. See the answer to Is there convection in Mars' mantle?
Second, tectonic activity is caused by internal heat. As a planet gradually loses its internal heat, tectonic activity slows down, until it eventually stops. There is no need for more convoluted explanations.


Could the Sun have liquid iron in orbit? How close of an orbit to the Sun could an iron meteor get before it melts into a ball?

As this source stats, galvanized building iron has an albedo of 0.351.

In general, we can assume, that a half of its total surface is shined by the Sun, with an average 45% angle. That means that it gets the radiation of the 6000 K Sun on $\frac{1}{2} \cdot \frac{1}{\sqrt{2}} \approx 35\%$, as if it would be shined directly on its whole surface. It radiates back 35% of it, the remaining 65% heats it. That means, in its equilibrial temperature, the iron meteor radiates $\approx 23\%$ in all direction due to its temperature, than it gets from the Sun.

Furthermore, being $n$ solar radii away from the Sun, it also gets the per-surface thermal energy of $\frac{1}{n^2}$. Here I committed yet another "approximation", because the Sun is not a pointlike radiator if $n$ is small, and the final result will be probably small, but it won't be too big.

The thermal radiation is proportional roughly with the 4th power of the temperature.2 Thus, $n$ solar radii away, it gets $\frac{1}{n^2}$ per-surface solar heating, it radiates away $0.23\frac{1}{n^2}$, what makes its equilibrial temperature $\sqrt[4]{0.23 \cdot \frac{1}{n^2}} = 0.69\frac{1}{\sqrt n}$ of the Sun's.

The average surface temperature of the Sun is 5778 K, the meltig point of iron is 1881K. Thus,

$$5778\ \mathrm K \cdot 0.69\frac{1}{\sqrt n} = 1881\ \mathrm K$$

$$\frac{1}{\sqrt n}=\frac{1881\ \mathrm K}{5778\ \mathrm K\cdot 0.69}=0.47$$

$$n=\frac{1}{0.47^2} \approx \underline{\underline {4.5}}$$

The answer is: 4.5 solar radius from the solar center, which is $\approx$ 1.8 solar diameter from the surface. It is far deeper than the orbit of the Mercury. As a comparison, the Parker Solar Probe will near the Sun 8 solar radii away. A 2000 K maximal temperature is predicted on its heat shield (note, that probe won't have an equilibrial temperature).

Another remark: liquid things in vacuum evaporate. The evaporation for molten metals near their melting point is slow, but it exists. Such an object, if it does not have something to keep the iron vapor around it (f.e., gravitation), will slowly disappear.

1Surprisingly, I did not found better data with google.

2Also this is not exact, because it is true only for absolute black body radiators, but the spectrum of most ordinary materials does not cause a huge error in the calculation.


What was the process of calculating how much darkmatter/matter/dark energy the observable universe consists of and how did it come at this conclusion In 2015 the Planck satellite consortium released an analysis of data collected by the Planck satellite that showed this percentage break down of the universe’s mass/energy.

Ordinary matter: 4.9%

Dark Matter: 26.8%,

Dark Energy: 68.3%

How did they calculate the "stuff" of the observable universe and how did they come to this conclusion?

I assume  this took some tough calculations and some work with a supercomputer but i wonder what the entire process was.

We are not actually measuring the $\Omega_r$,$\Omega_m$,$\Omega_{\Lambda}$.
At first, we measure the distance modulus to certain celestial objects by using the standart candles in the astronomy.
This distance can be calculated from,
$$d_L=\frac {c} {H_0}z(1+\frac {1-q_0} {2}z)\,\,\, (Eqn.1)$$
Here $q_0$ represents the deceleration parameter. Now I am not going to go in mathematical details but you should now that, using the acceleration equation we can write $q_0$ in terms of $\Omega_r$,$\Omega_m$,$\Omega_{\Lambda}$.
$$q_0=1/2 \sum \Omega_{w,0}(1+3w)$$  and for matter ,$w=0$, for radiation $w=1/3$ and for dark energt $w=-1$ we get,
$$q_0=\Omega_r+1/2\Omega_m - \Omega_{\Lambda}\,\,\, (Eqn.2)$$
And the relationship between distance modulus $(m-M)$ and $d_L$ can be written as,
$$m-M=5log_{10}(d_L/1Mpc)+25\,\,\, (Eqn.3)$$
If we combine Eqn.1 and Eqn.3 we get a relationship between distance modulus and z.
We plot the $m-M$ for distant objects with respect to $z$. From the slope of the graph we can determine the $H_0$ (for $z\rightarrow 0$) And for large z,  deviation of the plot from the straight line gives us to determine the $q_0$.
Now we have experimental data. Then we put different values for $q_0$ or equally $\Omega_m$ and $\Omega_{\Lambda}$ (with respect to Eqn.2) to determine the best fit for the graph (since we can take $\Omega_r=0$)

Edit: I clearly missed the "The Planck data" in the OP's question but its the same general idea. For different density parameters, the CMBR would look different, like the graphs that I shared are different for different density parameters. So by looking at the CMBR, we can determine the density parameters and which one fits well to the CMBR.

Further Note: I want to also suggest that the $\Omega_b$ can be calculated from the early universe by measuring the amount of deuterium and other elements present in primordial gas clouds.
And observering the rotational curves on the galaxies we can have approximate values on the $\Omega_{dm}$. (finding the dark matter density)
We cannot deduce the $\Omega_m$ and $\Omega_{\Lambda}$ just by looking at the Supernova data, for example the supernova data is also consistent with $\Omega_m=0$ and $\Omega_{\Lambda}=0.4$ or with $\Omega_m=1$ and $\Omega_{\Lambda}=1.7$


Why doesn't it hurt to look at stars? If I stare at the sun, it's bright and blinding. However, if I look at a star in the sky, it doesn't hurt. What causes this loss of energy that would otherwise damage one's eyes?

What's more, stars are so small to our eyes that they look like "point sources," meaning that each resolution element in our eyes gathers light from a much larger region of the sky than a distant star can fill.  If you were to look at the Sun from, say, Uranus, and you have good eyes, it would still hurt almost as much because you could still resolve the Sun at that distance (sharp eyes can resolve objects 1/25 the width of the Sun).  That means whatever spot on your optic nerve the sunlight is focused to would be getting the same heat that it gets now-- there would be fewer such spots if you were at Uranus, but a burning spot still hurts.

But any distance larger than 25 astronomical units or so, you would start to not be able to resolve the Sun so even that single spot on the optic nerve would be mixing in light from the dark region around the Sun.  At $10^4$ times farther away than Uranus, the distance mentioned in the other answer, it means the sunlight would get diluted by the darkness around the Sun $10^8$ times larger than the Sun, accordingly reducing the heat that spot on the optic nerve receives by $10^{-8}$.  Far enough away, you can't even see there is a star there.


Would bacteria on incoming meteors burn before impacting? I was watching an old Ancient Aliens last night and there was a whole episode dedicated to how the "Gods" (aliens) were spreading disease/plagues.

A good portion of the show was dedicated to how aliens may be sending meteors to Earth with bacteria to spread disease.

Part of this was several scientists talking about an experiment with a rock set in orbit (forgive me that I don't have a link to this).  Said rock had many forms of bacteria on it and after a certain amount of time they grabbed the rock and confirmed that some bacteria can survive space conditions.

Then there were other scientists saying that some forms of bacteria/virus may even get worse in zero gravity or space conditions - used salmonella as an example.

But my question is does this even matter?  Wouldn't any bacteria or virus colonies on a meteorite just burn up in our atmosphere?

Any bacteria present in a meteor would not necessarily burn.  According to the Bad Astronomer's "Bad Addendum" to another unrelated point, "A Meteoric Rise", meteors sure do get extremely hot, but that outside part usually ablates away, leaving the inner part that was extremely cold for who knows how long, which may not have warmed up much during the fall.

Bad Addendum: many people think that a meteorite, after it hits the ground, is very hot and glows red. Actually, meteorites found shortly after impact tend to be warm, but not hot at all! It turns out that it certainly is hot enough to glow while it is in the part of the atmosphere that decelerates it the strongest, but any part that actually melts will be blown off ("ablated") by the wind of its passage. That leaves only the warm part. Even more, the meteor is slowed down so strongly as it moves through the atmosphere that the impact speed is typically only a few hundred kilometers an hour at most. Only the very large (and we're talking meters across) meteors are still moving at thousands of kilometers an hour or more when they impact. Small ones aren't moving that fast at all. Not to say you'd want to be under one: a car in New York was struck by a small meteorite and had a hole punched through it, and the whole back end crushed in. Ouch!

This American Meteor Society resource seems also to indicate that meteorites can fall to the ground without being extremely hot.

Are meteorites “glowing” hot when they reach the ground?

Probably not. The ablation process, which occurs over the majority of the meteorite’s path, is a very efficient heat removal method, and was effectively copied for use during the early manned space flights for re-entry into the atmosphere. During the final free-fall portion of their flight, meteorites undergo very little frictional heating, and probably reach the ground at only slightly above ambient temperature.
For the obvious reason, however, exact data on meteorite impact temperatures is rather scarce and prone to hearsay. Therefore, we are only able to give you an educated guess based upon our current knowledge of these events.

This study, "Bacterial Spores Survive Simulated Meteorite Impact", seems to indicate that it's possible, though unlikely, for a bacterium to survive the pressure shock from such an impact.
So it's plausible that an extremophile bacterium could somehow survive this impact by being on the inside of the meteoroid, surviving the ablation process as a meteor, and not even get warmed up to the surrounding atmospheric temperature when it crashes as a meteorite.
There is nothing to indicate that this has actually happened.  But what a discovery it would be if something like this were to happen and be confirmed.


Two body orbit dynamics for orbital periods approaching light cone limit How do you model a system of two large celestial bodies, where they achieve a rotation very close to the limit at which information can escape the system (heat, light, etc.), do they begin to appear like toroids? Does the last parsec problem possibly discuss such things?

I am trying to take into consideration the object's radius as well.

As far as I know the only option here is numerical general relativity. There are no known exact solutions for the general two-body problem in GR, since the field equations are very non-linear. Even there there are limits, since when the fields get strong enough the bodies will start to deform in complex ways that also depend on the equation of state of the matter they are made from (and magnetic fields!). This is

However, current models of merging neutron stars do not usually produce toroids. The stars become elongated, splashing out matter that either forms tails or a disk. Some examples:

https://www.youtube.com/watch?v=Dyn9KbB_zeo
https://www.youtube.com/watch?v=AW_W-Fbrdu8&list=PLSYkic-CsfSGFb3KY2cjLA1EZyNDadK_w
https://www.youtube.com/watch?v=ow9JCXy1QdY

The reason for this is that maintaining a toroid against self-gravity is fairly hard, making it tend to elongate into an ellipse and then a spiral.


Question about Hubble parameter (Hubble constant) and measuring it I see this question in "An introduction to modern cosmology - Andrew Liddle - Wiley Publication":

In the real Universe the expansion is not completely uniform. Rather,
galaxies exhibit some random motion relative to the overall Hubble
expansion, known as their pecuLiar veLocity and caused by the
gravitational pull of their near neighbours. Supposing that a typical
(eg root mean square) galaxy peculiar velocity is 600 km/s, how far
away would a galaxy have to be before it could be used to determine
the Hubble constant to ten per cent accuracy, supposing
(a) The true
value of the Hubble constant is 100 km/s Mpc?
(b) The true value
of the Hubble constant is 50 km/sMpc ?
Assume in your
calculation that the galaxy distance and red-shift could be measured
exactly. Unfortunately, that is not true of real observations.

I really don't know how to start solving this. What do we need to find Hubble constant. I know $v=H.r$ but I don't know how to calculate the effect of peculiar velocity. we don't know $v$ and $r$ in Hubble formula, how to find $H$?

You can start by working out how fast a galaxy must be receding if its peculiar velocity (given as about 600 km/s) is less than or equal to 10% of its expansion velocity.

I'm not going to spell out the next step as you said you don't know where to start.


Can a telescope image be amplified to see more by adding distant lenses or mirrors? Could a lens or mirror be placed in orbit on the same flight path with other parts to make a telescope or on the horizon of the moon to see farther with the Hubble?

Speaking in general, telescopes are designed as complete systems from the beginning. It's not generally feasible to improve performance by just adding yet another lens or mirror to the assembly. The reasons for this are not simple to explain, since this is basically the whole theory of instrument design.

Usually you start with a set of requirements (limiting magnitude, angular resolution, linear resolution in prime focus, etc) and with a set of constraints (cost, size, weight, materials).

You settle on a given architecture (newtonian reflector, Ritchey-Chrétien, doublet / triplet / quadruplet refractor, etc), and then start to work out the main parameters: aperture, focal length, etc.

There's an iterative process involving computer aided design to work out the details of each optical element; at each step you need to solve problems such as eliminating internal reflections, etc.

In the end, what you have is the design of the whole optical stack: say a Dall-Kirkham reflector, 1 meter aperture, f/8, with mirrors of whatever radiuses of curvature, etc.

Once that's set in stone, it's hard to change in a major way. Minor upgrades might be possible (add a corrector lens, or a different secondary mirror), but not huge modifications.

TLDR: No. Start with a bigger instrument from scratch if that's what you want.


Why are the planets moving? My question isn't why it doesn't have a circular orbit, and why an elliptical, my question is why do they move at all? What caused the debris in space to apply such a force on the planets that it has been circling around the Sun billions of years.

This is a simple explanation to give a basic understanding as to why the solar system rotates.

It's caused by gravity and velocity. If you put two objects with mass in space a distance apart and at rest, they will accelerate directly toward each other, until kerpow! they hit.

But, if they have some small, random relative velocity to begin with that's not directly toward or away from each other, then no amount of acceleration toward each other will zero that out. It means they will pass each other at some finite distance and keep going. If the random velocity is low enough, this dance will repeat and they will move in elliptical orbits around their center of mass.

This kind of motion can be expresses in terms of its angular momentum and in this example (as in most) it is going to be a conserved quantity, so it's always going to be there.

So when things in the universe like solar systems or galaxies initially form, they have some random, residual, collective motion, and this angular momentum in the center of mass frame of whatever blob is coalescing doesn't go away. Objects in the solar system keep rotating because there's no other option, at least in the short term.

If there was a large amount of gas in the solar system, that might slow things down a bit, and there is a teeny tiny loss of energy due to gravitational waves, although for planets this is so extremely small it can be ignored. But basically the "wheels" of the celestial sphere are well lubricated and solar system bodies will continue in their orbits for millions and perhaps billions of years because there's nothing we know about right now to stop them.


What is meant by *topocentric* right ascension and declination? I'm looking up how to convert an azimuth/elevation measurement from a specific location on Earth's surface to right ascension/declination.  I'm coming across the terms "topocentric declination" and "topocentric right ascension", which I haven't come across before.

What does the "topocentric" mean here? Is this different from "regular" RA and dec?

My guess from the terminology is that, for objects that can't be treated as being infinitely far away from Earth, "topocentric" is used to differentiate a sky position of an object as seen from a specific place on Earth instead of from Earth's center.  (For very distant objects, like stars, the sky position seen from Earth's surface and Earth's center are for virtually all purposes going to be the same.)

If my guess is correct, is there a convention specifying which equator and equinox the topocentric RA and declination use?  Is it supposed to always be true equator and true equinox, or something more along the lines of a J2000 epoch?

You are correct that topocentric coordinates are for the position of "close" objects, corrected for observing from the Earth's surface instead of the theoretical center.

The topocentric and geocentric coordinates would use the same system (J2000, true, etc.) There is no convention to use one in favor of another (but there may be preferences in some cases).


Matter/mass exist into the space. Where does space exist? It is said that before big bang there was nothing. Not even space/time. So, now where does space exists?

There is still no way to solve this question for all people.
You also need to keep in mind that time is directly linked to space. So before the universe there was no space and no time. However, if you want describe processes or even states you should also be able to talk about time.
Time is the perception of a sequence of events. So before the big bang no events could happen. To be more detailed time is what physicsists call the increase of entropy. Entropy itself is the randomness of a closed physical system. Each physical systems wants to have maximum entropy or randomness. The big bang at point of time -zero- can be seen as a point with the lowest possible value of entropy, because any matter/energy was located at a single point - at a singularity. There was ne place for particles to get at a random place.

With the expansion of this point space and time expanded into "nothing". If you think about it: how could it expand if there was something?

Next: If there is something outside, we still have a neverending question: What's outside the outside? Is it infinite?
Our physical laws don't work outside - so how should we talk about things nobody could ever describe?

Of course, there are theories that try to explain why or how the big bang happened - the ethereal expansion by Alan H. Guth and Alexander Vilenkin, for example. Universes are "created" like holes in a cheese by massive discharges of energy (like lightnings during a storm), but those are very hypothetical theories without any prove or even evidence. And all those theories need parallel universes / multiverses.

So, for my opinion, I keep at this point: There was nothing. As human beings we cannot deal with nothing. For us there is always something. If we try to think about nothing, we still think on how much time has passed not thinking about anything - for example. Nothing is mathematically described as 0. But also 0 is a number. It's more than nothing. It still consumes place in a formula. So don't think about nothing as "nothing". Think about nothing as everything. Because or universe was created out of nothing, this nothing is also everything for any or every time.


Why does Titan have lower surface gravity than the Moon when Titan is more massive? Reading on Wikipedia I saw that Titan is 80% more massive than the earth's moon but has only 85% the surface gravity. Why is this?

Surface gravitational acceleration on an object with mass $M$ and radius $R$ is given by
$$
g = \frac{GM}{R^2} \propto G\rho R
$$
where $\rho \propto M/R^3$ is the density of the object.
If one body has smaller surface $g$ than another, it must have smaller density $\rho$, smaller radius $R$, or both.
Titan is larger than Earth's Moon, so your observation about its surface gravity means Titan must be less dense than the Moon.  Wikipedia confirms:

$R_\text{Titan} = 1.5 R_\text{Moon}$, but
$\rho_\text{Moon} = 3.34\rm\,g/cm^3$ while Titan has only $\rho_\text{Titan} = 1.88\rm\,g/cm^3$.


What's the relation between Einstein's Gravitational Theory and Dark Matter? I was reading an elementary book on dark matter (in fact, a historical perspective) and there were mentioned how the scientific community react to the idea of dark matter proposed as a solution to observed discrepancy between the actual mass of astronomical systems and the predicted mass from Newton's theory. I was wondering where Einstein theory stands in relation to dark matter, did it somehow predict it, or does dark matter prove the incompleteness of Einstein's theory? And what about dark energy?

Dark matter was originally hypothesised because there is a greater degree of rotational energy in galaxies than visible matter would allow for - crudely they rotate so fast that they ought to spin apart and therefore it was hypothesised that there is an additional source of gravitational force in some form of invisible matter.

This is not as a result of a discrepancy in Einstein's General Relativity. Newtonian gravitation is a very good approximation of Einstein's GR at most ordinary scales and energies and this is true in this case also.

I am not sure, exactly, what your "what about dark energy" question means, but assuming you mean, does this imply an incompleteness in GR, then again the answer is no, or at least not necessarily, but it's a more subtle point.

Dark energy is the hypothesised source of the energy that causes the universe to expand at an accelerating rate. It can be inserted as a term into the field equations of GR (other explanations of DE exist though) - but that is just a mathematical term, rather than an explanation of the physicality of it.

Einstein originally inserted such a term into his solutions - to predict an essentially static universe. When it was shown that the universe was expanding he described this as his "greatest mistake".

The evidence for the accelerating rate of expansion is relatively recent and came long after Einstein's death so he never lived to see the essential idea - of a "cosmological constant" in his equations - revived.

This explains more - https://en.wikipedia.org/wiki/Cosmological_constant


Temperature of a substellar point on an airless tidally-locked planet If a hypothetical blackbody planet's components were solid at all temperatures, would the substellar point on the airless tidally-locked (1:1) planet eventually heat up to the surface temperature of the star it is orbiting? This material would be a very poor heat conductor.

Also, would the temperature on the opposite point approach the cosmic background temperature 2.7 K?

The planet would reach an equilibrium where the amount of heat absorbed is the same as the amount of heat radiated. If there is no way to transfer heat on the planet (no conduction, no atmosphere), then that condition must apply locally.

The flux radiated from a blackbody surface (in W/m$^2$) is given by $\sigma T^4$, where $\sigma$ is Stefan's constant and $T$ is the temperature.

If the substellar point is a distance $d$ from the star, and the luminosity of the star is given by $L \simeq 4\pi R^2 \sigma T_{*}^4$ (assuming it too is a blackbody with radius $R$ and temperature $T_*$) and assuming $d \gg R$ to avoid some unnecessary geometric unpleasantness, then the flux absorbed at the substellar point is $L/4\pi d^2$. The flux is all absorbed, since you wish to assume a blackbody.

The temperature of the substellar point at equilibrium is therefore given by
$$\sigma T^4 = \frac{4\pi R^2 \sigma T_*^{4}}{4\pi d^2},$$
$$ T = T_* \left(\frac{R}{d}\right)^{1/2}$$

Since we assume that $d \gg R$ then clearly $T < T_{*}$.

At other points on the planet's surface it will receive a reduced flux from the star, simply becasue the flux from the star is incident at an angle to the exposed surface, so the equilibrium temperature will be lower.

On the unlit side of the planet there is no illumination from the star, but an almost isotropic flux from the cosmic microwave background equal to $\sigma T_{\rm CMB}^4$ over the entre surface. Therefore in the absence of any other source of heat, then that side will assume the temperature of the CMB at equilibrium.


Did Einstein supported Big Bang Theory cosmological model? Einstein made many predictions, including gravitational waves and the possibility of black holes.

Relativity is taken into consideration for the Big Bang model, so did Einstein agree with it or did he have a different view?

Einstein's opinions were not static, and he lived at a time when there were several competing theories and not much observational evidence.

Einstein introduced a cosmological constant $\Lambda$ into his equations, the purpose of which was to allow for a steady cosmos with no expansion or contraction. He is said to have later regretted this addition. He was aware of Hubble's observations of galatic redshift, and Lemaître's "hypothesis of the primaeval atom". But along with most physicists in the 1920s he did not support it. He suggested an oscillating model in the 1930s, but rejected it later (apparently because it didn't solve the problem of entropy)

His final opinion then seems to have been "we don't know".  When he died in 1955 there was no consensus, and Einstein was by then an old man and not actively involved in research physics.


How can we predict a Big Crunch when all galaxies are moving further apart? I've read that the further away a Galaxy is from us the faster it moves away.

By this logic how can scientists predict that there will eventually be a big crunch when every piece of matter is seemingly getting further and further apart?

In a homogeneous and isotropic Universe (even if recent observations challenge this hypothesis), you can derive the Friedmann equations, which describe the evolution of the Hubble constant with time:
$\frac{\dot{a}}{a} = H(t) = \frac{8 \pi G}{3}\rho - \frac{k}{a^2} + \frac{\Lambda}{3}$   (with $c=1$) (Equation $1$)

where $a=a(t)$ is the scale factor, $\dot{a}$ its derivative, $G$ the gravitational constant, $\rho$ the matter density, $\frac{k}{a^2}$ the spatial curvature (a parameter that describes the metric of the Universe), and $\Lambda$ the cosmological constant (an integration constant added by Einstein).
It could be useful to rewrite the equation as:

$H^2 = \frac{8 \pi G}{3}(\rho + \rho_{\Lambda}) - \frac{k}{a^2}$

where $\rho_{\Lambda} = \frac{\Lambda}{8 \pi G}$ is the "density of cosmological constant".

We can also expand the matter density as $\rho = \rho_{matter} + \rho_{radiation}$.

So we have a "total" density $\rho_{tot} = \rho_{matter} + \rho_{radiation} + \rho_{\Lambda}$. The destiny of the Universe depends on this amount.

In case of $\rho_{tot} > \rho_{crit}$, or equivalently a closed Universe ($k=+1$), the equation $(1)$ becomes:

$\dot a^2 = \frac{8 \pi G}{3}\rho a^2 -1$

Which points out that the scale factor must have an upper limit $a_{max}$ ($\dot a^2$ must be positive).
This in turn means that the second derivative $\ddot a$ of the scale factor must be negative, when approaching $a_{max}$, that is the scale factor function inverts its behavior:

Look at here and here if you want to go deeper.

@Bardathehobo This figure shows what I mean when I say that a currently accelerating Universe can still crunch. This is because we are basically ignorant upon the dark energy issue.


Why does the Moon sometimes look Yellow? My friend and I both noticed that the Moon looked yellow Yesterday, but it wasn't like that the same time the day before, as far as I remember.

Does the Moon sometimes look Yellow? If it does, why does it?

Because light with higher frequency, that is bluer colors, scatters more in Earth's atmosphere and celestial objects thus look redder (yellow is towards red) while the atmosphere looks blue. See Rayleigh scattering.
Here is video with a lengthy and somewhat entertaining explanation:
https://www.youtube.com/watch?v=SRh75B5iotI


Mountains on the moon during solar eclipse? I ask myself the reason of why the sun in not uniform on the edge of the moon at this moment of the eclipse (cf. picture) is not unifom is because of the mountains of the moon ? Or can it due to erruptions of sun ? If not is there another reason to explain that ?

The beads of bright light can be seen thanks to the Moons rugged topography. They're called Baily's beads in honor of an English astronomer who explained the phenomenon, Francis Baily.

In the photograph you posted, you can also see Solar prominences. It's plasma ejected into the corona. Its temperature is lower than that of the coronal plasma and it emits more visible light.


Estimates of exoplanets distribution consistent with current data I am interested in current estimate of distribution of planets of various radii in various distances from their parent star. There is many sources, where one can find database of presently known exoplanets, for example exoplanets.org, www.openexoplanetcatalogue.com and others. However, the known exoplanets are, of course, subject to conditional probability of detection by various methods. One cannot take the database as representative distribution, because, for example, hot Jupiters or big planets in general would be greatly overestimated.

What are the best resources to obtain current unbiased estimates of exoplanet radii and distances?

As you've correctly mentioned observation biases play a huge role in current understanding of planetary populations. Let's have a look at a slightly outdated plot (age 2 months) that shows us the semimajor-axis vs mass distribution of currently known planets including KOI's (unconfirmed planets).

Here we see two things:

Hot Jupiters seem to make up a significant part of the confirmed planets (confirmed = RV + Transit)
The farther out we go, the less likely it is to have a measured planetary radius (black dots)

Now knowing how our methods and instrumentation work, we can try and debias this data.
Transit measurments pose the easiest ones to do this: as for transits the detection probability is a simple geometric function of planetary alignment, radius and stellar radius one can overplot the found planets with this probability. So this is usually of not very massive interest to the planetary community.
edit
The probability usually is given as $\sim r_{stellar}/2 a$, with $r_{stellar}$ being the stellar radius and $a$ the planet's semi-major axis. A simple derivation is here. Discussion how everything gets more complicated when combining with RV measurments here.
/edit

Any instrument one uses however, will have their own quirks and biases and starspots or eclipsing binaries can lead to false positives. Petigura+ 2013, fig. 1 try to asses those effects for Kepler transits and come up with the following detection probability map.

The corrected population then would be found by dividing $\rm\frac{found population}{completeness}$. Mathematical issues like dividing by zero are easily interpreted as issues stemming from the incomplete data. As where the detectability is zero, we can't know if there are 0 or 1 or 20 planets hidden in a bin.
When projected on one axis or the other those corrected data-plots then result in corrected Radius- or Period occurence rates.
I won't show those for the transit method, as I want to encourage to check the original reference.

I show one of those plots for the radial velocity method however. From Mayor+ 2011, fig. 12 that analyse their HARPS-spectrograph data from the preceeding decade:

In black: Found planets, red: debiased radial-velocity population
Note: A Jupiter-mass is roughly 320 Earth-masses, so in this plot the Hot Jupiters are on the far right. This now closes the thematic circle to my initial note about Hot Jupiters: While in black Jupiter-like planets seem very common, in the debiased version this is not the case anymore. Mayor et al. estimate the Hot Jupiters to make up only 0.5% of the total planetary population with this method.

Note also that although the debiased number of terrestrial planets increases significantly, their error bars explode in size. This stems simply from doing statistics with low numbers. Also for reasons of simplicity and clarity the error bars on the first two plots are not shown.

Concluding:
It is important to debias the data we have about exoplanets by method.
When doing this, we get rough ideas about how the true planetary populations look like.

Recent results and non-stop public announcments about "hundreds of newly found planets" can lead to a certain, totally understandable euphoria that we now know everything about planets.
This is however very far from true. The Number distribution in full mass-radius-semimajor axis space is still very, very thinly spread. So in the end we do low-number statistics when debiasing, which is scientifically simply unsatisfying.
Or said differently: There is still a lot of exciting work to be done!


Could an asteroid impact cause radioactive fallout? Could a (regional or global) fallout of radioactive material be a "bonus" disaster effect of an asteroid impact? My reasoning for how such a scenario maybe could be possible:

A) Some asteroids maybe consist of lots of radioactive heavy metals because they were initially formed by ejecta from the cores of planets during planetary collisions. Maybe even an airburst of a rather small such an asteroid could cause widespread dangerous radiak fallout?

As I understand it, Earth's initial uranium has fallen into the core. The uranium mines dig into concentrated deposits of uranium in the crust which came from uranium rich asteroids.

B) If any kind of asteroid large enough hits a uranium rich part of the Earth crust (making it the second one on the same spot), it could maybe eject much of Earth's own uranium and thus cause a dangerous fallout?

I'm not aware of any relevant uranium ore deposit, which is related to meteoritic material.

Dangerous fallout is by far the most caused by short-lived radioactive isotopes. Those isotopes are rare in meteorites as well as in rock on Earth.

Natural nuclear reactors, which would produce short-lived radioactive isotopes, don't occur any more on Earth, since the natural U-235/U-238 isotope ratio is too low, now. This is due to the half-life of U-235 of about 713 million years in contrast to 4,468 million years for U-238.

If an asteroid would impact into a uranium ore deposit, most of the molten ejected material would fall back to Earth as tektites.

Evaporated ejecta and aerosols (dust) would probably increase health risk for some time, until it's washed out from the atmosphere. But other, less-radioactive aerosols would be a health risk, too.


What is "class 4 light pollution" and could I estimate my own area's class visually? This question currently asks

  I am living in the northern hemisphere in Greece and in class 4 area (light pollution scale).Is it possible to see the Andromeda galaxy with a naked eye?

I'd never heard of a light pollution scale before. How does the scale work, and if I wanted to gauge or estimate my own area, can I do that by looking at objects in they sky visually?

Only the original poster can answer what was meant, but one system for rating the darkness of the night sky is the Bortle Scale.

https://en.m.wikipedia.org/wiki/Bortle_scale


How do I plot galactic coordinates using matplotlib and astropy in python? I want to make a projection of millisecond pulsars on the galactic plane, much like this one from Sala et al. 2004:

I have tried several methods, and have not gotten anywhere. This is my current code, along with what it is producing:

import math as m
import numpy as np
import csv
import matplotlib.pyplot as plt
import pandas as pd
import astropy.coordinates as coord
import astropy.units as u
from astropy.io import ascii
from astropy.coordinates import SkyCoord

data = pd.read_csv('galacticwperiod.csv')

xarr = np.array(data.iloc[:,0])
yarr = np.array(data.iloc[:,1])
eq = SkyCoord(xarr[:], yarr[:], unit=u.deg)
gal = eq.galactic

#print(xarr)
#xarr = np.array(df.iloc[:,0])
#yarr = np.array(df.iloc[:,1])
#zarr = np.array(df.iloc[:,2])

#ra = coord.Angle(xarr[:], unit=u.hour)
#ra.degree
#ra = ra.wrap_at(180*u.degree)
#dec = coord.Angle(yarr[:], unit=u.deg)
#print(ra)

plt.figure(figsize=(6,5))
fig = plt.figure()
ax = fig.add_subplot(111, projection="aitoff")
plt.plot(gal.l.wrap_at(180*u.deg), gal.b.wrap_at(180*u.deg), linestyle='None')
ax.scatter(gal.l, gal.b, linestyle='None')

#ax.set_facecolor('xkcd:battleship grey')
#fig.patch.set_facecolor('white')
#ax.tick_params(axis='both', which='major', labelsize=10)
#ax.grid(color='b', linestyle='solid')
fig.show()
#plt.savefig('millisecondcoloraitoff.png', dpi=600)

Here are a few lines of the input file 'galacticwperiod.csv':

Gl,Gb
111.383,-54.849
305.898,-44.888
305.913,-44.877

This is the image it produces:

I am almost certain this is wrong because they are not distributed along the galactic plane, which they should be. The data I am using is from the ATNF Pulsar Catalog.

These are the sites I have already looked at, for reference:

https://stackoverflow.com/questions/33105898/astropy-matplotlib-and-plot-galactic-coordinates
https://astropy4scipy2014.readthedocs.io/_static/notebooks/06_Celestial_Coordinates_solutions.html

Any help with this would be greatly appreciated.

This code reads coordinates as equatorial (ra, dec) and transforms them to galactic (l, b):

eq = SkyCoord(xarr[:], yarr[:], unit=u.deg)
gal = eq.galactic

The contents of 'galacticwperiod.csv' are already in galactic coordinates and should not be transformed.
Something like this may give better results:

gal = SkyCoord(xarr[:], yarr[:], frame='galactic', unit=u.deg)

The other issue is that pyplot's geographic projections seem to expect angles in radians.
This code:

plt.subplot(111, projection='aitoff')
plt.grid(True)
plt.scatter(gal.l.wrap_at('180d').radian, gal.b.radian)

produces this plot:


Why don't the stars in a binary star system of equal masses always orbit their center of mass in a circular orbit? If I have a binary object system of equal masses, why don't they always have the same circular orbit around their center of mass, like in the photo?

There's no reason why 2 bodies of equal mass couldn't have elliptical orbits around each other.   There's an example of that here

.

The simple way to think about this is, if two bodies of similar mass approach each other, one of two things can happen, they either have sufficient velocity to pass each other with some hyperbolic curving of both objects or there's enough gravitational attraction that they capture each other in orbit, but the shape or eccentricity of the orbit depends on the ratio between the tangential velocity and the orbital speed at the closest pass.   For a circular capture the ratio needs to be exactly 1, and that kind of exactness is rare, so we nearly always get an elliptical orbit with an orbital capture.

Solar-system planet or moon formation (not moon capture) tends to be much more circular, because elliptical orbits of objects tends to cancel out as large bodies coalesce, but you're never going to get 2 similar mass objects orbiting each other in a condensing gas cloud.   Most of the mass inevitably collects in the gravitational center.  There can be, however, a tidal influence that circularizes orbits over time, never reaching a full circle, cause that's impossible, but slowly becoming more circular.


What galaxy looks the most like the Milky Way? Is there a particular galaxy cataloged by astronomers which is considered to most closely resemble what the structure of the Milky Way is believed to look like, when viewed from a similar distance?

According to this article on space.com the most similar looking galaxy to our own is NGC 6744. The source of this article comes from a statement by the ESO (European Southern Observatory) which says:

If we had the technology to escape the Milky Way and could look down on it from intergalactic space, this view is close to the one we would see — striking spiral arms wrapping around a dense, elongated nucleus and a dusty disc.
There is even a distorted companion galaxy — NGC 6744A, seen here as a smudge to the lower right of NGC 6744, which is reminiscent of one of the Milky Way’s neighboring Magellanic Clouds

Included below is an image of the galaxy (NGC 6744) in question, taken with the Wide Field Imager on the MPG/ESO 2.2-meter telescope.

However it is also noted that while the Milky Way and NGC 6744 are similar in appearance, NGC 6744 is almost twice the diameter of the Milky Way.


Rotation of the Earth about its own axis I am wondering how the Earth got its original angular velocity. What was the cause that made the Earth rotate about its own axis?

What about its motion around the Sun? What gave the Earth the push to keep rotating around the Sun?

Origin of Earth rotation
There are two main phenomena that explain Earth rotation. First one is linked to the history of the solar system: there is angular momentum in the interstellar medium (ISM), so the collapsing cloud that formed our solar system had an initial angular momentum (rotation of the galaxy and turbulence in the ISM are typically a good candidate for that; see for example Mac Low & Klessen 2004). By conservation of the latter, particles in the collapsing cloud accelerate (the norm of the angular momentum $I$ is $mrv$, with $m$ the mass of the system, $r$ the radius and $v$ the velocity; as the cloud contracts, $r$ decreases and then $v$ has to increase) and due to centrifugal force, a disk will tend to form. By conservation of the angular momentum in the protoplanetary disk, you will also tend to form bodies that rotates.
The second phenomena is linked to the formation history of the Earth. Simply put, Earth formation history is due to collisions of planetesimals (let say they are big chunks of rock, formed by a collision history of smaller chunks of rock, formed by...); these planetesimals are orbiting the forming star in the protoplanetary disk, their rotation being linked to the disk formation history (see above). When planetesimals collide, two things can basically happen: either the collision is perfectly central (meaning that the momentum of one of the two colliding objects is pointed toward the center of mass of the second object) and there is no transfer of angular momentum, either the collision is lateral and there is a transfer of angular momentum. Depending on the configuration, you can either transfer momentum that will make the other body rotating on way or the other. That being said, you can then object that there should be a equirepartition of collisions pushing the proto-Earth to rotate in both directions. Which is true, but external planetesimals are rotating faster than internal ones. Therefore, from the proto-Earth perspective (meaning, in a physical sense, in its reference frame), internal planetesimals are seen as coming backward, from front, and external planetesimals are seen as coming forward, from behind. Long story short, there are both transfering momentum in the same "direction", forcing the proto-Earth to rotate in the same direction as the rotation of the protoplanetary disk (see for example Artem'ev & Radzievskii 1965).
Rotation around the Sun
This one's easy: gravity.
The Earth is constantly falling towards the Sun because of gravity, and the centripetal acceleration counteracts perfectly gravity, mainting the Earth on its orbit.


What is the difference between emission line and recombination line? When an electron jumps from higher level to lower level a photon is emitted. This is seen as an emission line. But what exactly is a recombination line? I found them similar. Can you please tell me what's the difference between these two?

A recombination line is a special case of an emission line.

Emission lines

An emission line is any spectral feature that rises above the continuum — i.e. the average amplitude of the spectrum (in some wavelength region) — and is due to atomic transitions (where "atomic" include atoms in molecules and dust grains, and "transitions" may be electronic, vibrational, or rotational).

These transitions may arise from absorption of light, or collisions of an atom with another particle. In astrophysics, this other particle will usually be a free electron, since these are much faster than atoms. The absorption of energy puts the atom in an excited state, and when the atom de-excites, a photon is emitted.

Recombination lines

Recombination is the process of a free electron being "caught" by an atom missing an electron, i.e. an ion. The ion, in turn, has been produced previously by being ionized by some high-energy process, e.g. an ionizing photon or a violent collision. The difference between the kinetic energy of the electron before the encounter, and the energy of the state it goes to, is emitted as a photon.

The electron may go directly to the ground state, in which case another iozing photon is just produced. But it may also go to an intermediate state, emitting a lower-energy photon. The electron is then in an excited state, from which it will soon de-excite, possibly through several levels until it reaches the ground state. This process is called cascading.

The photons that are emitted during this cascade are called recombination lines.

Recombination lines from gas around stars

In the vicinity of very hot stars (O and B stars) which produce many ionizing photons (the Lyman continuum; LyC), you will have a region of neutral hydrogen gas (HI) being ionized. Timescales for recombination in these regions are quite small, so the LyC is "almost immediately" reprocessed into recombination lines.

Quantum mechanics tells you the probabilities of ending up in the various states, and it turns out that, for instance, for each LyC photon you get roughly 0.68 Lyman $\alpha$ photons (with a small dependence on temperature), i.e. photons emitted when the electron false form the first excited to the ground state. In terms of energy, this translates into ~1/3 of the total power.

This is actually quite amazing: It means that 1/3 of the total continuum of photons more energetic than the 13.6 eV needed to ionize hydrogen is converted into a single emission line! For this reason, galaxies — in particular young galaxies that still host many hot stars — are often quite luminous in Lyman $\alpha$; sometimes even only in Lyman $\alpha$.

Absorption lines

Emission lines are related to absorption lines: In the first case, you have a continuum of light (e.g. from thermal processes) with some physical process "adding" some extra light to the total spectrum. In the latter case, you have the same physical process removing light from a continuum.

For instance, if you observe a (non-ionizing) source through an HI cloud, you'll see absorption lines at the wavelengths corresponding to the various transitions of neutral hydrogen. If you were able to observe the same cloud from another angle, you'd see emission lines at the same wavelengths.


Does the lower mass of Uranus suggest it swapped position with Neptune? As Uranus is less massive than Neptune, would this be evidence that it was formed further away from the sun than Neptune, and later it swaps orbit with Neptune to move closer to sun?

The Nice model of solar system evolution proposes that both Uranus and Neptune migrated outward from their original orbits due to numerous encounters with planetesimals,
stimulated by a temporary 2:1 resonance between Jupiter and Saturn.
Simulations going back billions of years can put either Uranus or Neptune initially closer to the Sun.
Desch 2007, assuming that the Nice model is correct, argues for a particular density profile of the protoplanetary disk and notes that a closer proto-Neptune fits that profile better.
Some solar system scientists have problems with the Desch model, but Uranus's smaller mass is consistent with it.


Determine Julian Date from Gregorian without formula As an exercise, I am trying to calculate the JD from a given date in the Gregorian Calendar at a time given in UT. Furthermore, I want to do so from first principles without relying on any formula. Let's say I choose UT 18:30 on 2/28/2010. I am not able to obtain a result consistent with the U.S. Naval Observatory results. I am in need of a calendar guru to show me where my calculation is going astray.

First I find the number of years from 4713 BCE (Julian Date 0) and the target year making sure to add an extra one so as to include 0 CE. This is $4713 + 1 + 2010 = 6724$ years. The number of these years which are leap is $\frac{6724}{4} - 3 = 1681$. This equation attempts to reflect that every fourth year is a leap year except for those years falling after 1582 (the year of the Gregorian Calendar reform) which are both divisible by $100$ and not by $400$. There are three such years which need to be deducted from the total-- 1700, 1800, 1900. This leaves $6724 - 1681 = 5043$ non-leap years.

Now I convert these to days and subtract the $10$ days which were skipped as part of the reform from October 4 to October 15 in 1582. $5043*(365) + 1681*(366) - 10 = 2,455,928$. Next, I find the number of days from January 1st to the target day in the target year (February 28th). This is $59$ days and so I have $2,455,987$.

Finally, the time number of seconds afternoon on the target day is calculated as a fraction of the number of seconds in a 24 hour period. This works out to be $\frac{6(60)(60) + 30(60)}{24(60)(60)} \approx 0.27$. Adding this to the earlier result yields $JD = 2455987.27$.

The actual answer is $2455255.77$ which leaves a difference of $2455987.27 - 2455255.77 = 731.5$ days. This is nearly two years. Obviously, I have either misunderstood the calculation or have performed it incorrectly. Any help would be much appreciated!

The Year 0 does not exist in BCE/CE and BC/AD notation. It jumps from 1 BCE to 1 CE directly. So if we look at the sequence of years in BCE/CE notation, we have:

3 BCE
2 BCE
1 BCE
1 CE
2 CE
3 CE

If you want to calculate the number of years between, say, January 1, 2 BCE and January 1, 2 CE, you can add the year numbers to get 4, but then you need to subtract 1 to remove year 0 from the total, giving a total of 3 years. In your particular case, this leads to $4713 - 1 + 2010 = 6722$ years.

In astronomy, we often use another year numbering system called the Astronomical year numbering, which includes a year 0 and makes things more convenient arithmetically when calculating time intervals between eras. It goes (-3, -2, -1, 0, 1, 2, etc.) To convert years BCE to the Astronomical year notation, we subtract 1 and add minus(-) in front of the year. For example, the year 4713 BCE is -4712 in Astronomical year notation. Your year interval is then calculated as $2010 - (\text{-}4712) = 2010 + 4712 = 6722$.

For the leap years calculation, since -4712 is a leap year, this simplifies matters, but don't forget to take the smallest integer greater than or equal to your result (the ceiling function). So in this case $\frac{6722}{4} - 3 = 1677.5$, and the ceiling is 1678, and that is the number of leap years between your years. Also, in your initial calculation, you had forgotten to subtract 3, because 6724 / 4 - 3 = 1678, not 1681.

Finally, the time interval between January 1 and February 28 is 58 days, not 59, because we do not count the last date as part of the interval. This gives a total of $5044 * 365 + 1678 * 366 - 10 + 58 + \frac{6 * 60 * 60 + 30 * 60}{24 * 60 * 60} = 2455256.27$.

You state that the actual answer is 2455255.77, but in fact the answer really is 2455256.27. Perhaps you typed "06:30" instead of "18:30" in the converter.


What makes the moon, a super moon Caught a sight of moon tonight (by chance) and came in surprise that moon looks thrice in size to its usual size.

Mostly called as Blue moon. Didn't succeed in getting the details and explanations of the phenomenon on Internet. Wikipedia just tells about the origin of the word blue moon and particular dates at which super moon is more obvious but doesn't shed the light on the true physical causes.

Just wondering what could be the causes that makes the appearance of moon (near the horizon) so big.

A Supermoon is when the point in the Moons orbit when it is at its closest to the Earth coincides with full or new Moon (see Wikipedia entry).

A blue Moon is (in this instance) the occurrence of a second full Moon in the same calendar month (see Wikipedia entry), and is unrelated to the coincident Supermoon.

Also the apparent size difference between a full Supermoon and a more usual full Moon is inlikely to be detectable by a casual observer. As can be seen in the side-by-side comparison graphic on the Wikipedia page reproduced below:

The apparent giant Moon you report is the well known Moon illusion where the Moon appears larger when observed near the horizon compared to when seen higher in the sky. The Moon illusion is startling if you are unfamiliar with it and very much more noticeable than a Supermoon to the casual observer.

While out in the morning of the 31st, walking the dog in the local recreation ground, between ~04:00 and 05:00 BST (03:00-04:00 UT) I observed the apparent size of the Moon and commented to the dog that we will probably see someone (in the media I thought) mistake the apparent size illusion I was seeing for the effect of the Supermoon. (I had also noted the same effect the day before at a similar time, but not on the 1st as we go out later at the weekend)


When will solar eclipses become lunar transits? When will the moon be far enough away from the Earth that we would no longer consider it a 'solar eclipse', but rather a 'lunar transit' when the moon travels between the Earth and Sun?

When looking at definitions of [transit] and [eclipse] or more general [occultations] it is noteworthy that the degree of covering is not defined, so technically speaking each eclipse is also a transit and an annular eclipse even more so.

Details for next eclipses can be found at [this NASA page]. The next annular eclipse, err, transit happens next Sunday 2017-02-26 and is visible in South America, south Atlantic and Afrika.

As you were referring to the [increasing distance between earth and moon] (+3.8 cm /yr), it then depends on an exact definition on when an eclipse is not regarded as an eclipse anymore. Assume for simplicity that this is the case, when the diameter of the moon has halved (one quarter the area), this means that the distance had to be doubled.
This would take [380 Gm] / 3.8 cm/yr = 10 Tyr (using SI prefixes).

Note that one might need to take other effects into account, too, for example the [increasing distance of earth and sun] (~15 cm / a) and that the [sun is becoming a red giant] in a couple of Gyr.

Sorry, would have loved to post more links, but are not allowed to currently.


For stars around 13th magnitude, are their entries in various catalogues usually linked, or does one do this manually? @MikeG's speedy answer to "What star was the asteroid 2014 MU69 predicted to occult on 2018-08-04?" lists three names for one 13th magnitude star, from three catalogues; 2MASS, UCAC4, and Gaia DR2.

My question could be slightly subjective, but I believe that someone familliar with the topic will be able to provide an authoritative answer.

For stars around 13th magnitude that do appear in all three, are they electronically linked between catalogs in some automated way, or do you pretty much check each of them, decide the objects are the same, and then make note of your conclusions if you are writing something up?

2MASS All-Sky Catalog of Point Sources (Cutri+ 2003)
UCAC4 Catalogue (Zacharias+, 2012)
Gaia DR2 (Gaia Collaboration, 2018)

If you search by position in the CDS Simbad database by position it will give you entries in all three of these catalogues. If you search by name, then it will look for matched objects in these three catalogues. I am unsure what the matching algorithm is - it must cross-correlation on the position and look for objects within an appropriate matching radius, taking account of the proper motion of the object and epochs of the various catalogues.

This will not be foolproof, but the density of 13th magnitude objects on the sky is low enough that false matches are highly unlikely.


Seeking advice regarding buying a telescope I'd like to buy a telescope with a focal length of 700mm and an aperture of 60mm.

Would I be able to use a telescope of this power to study the closer planets (for example Mars and Venus) and nebulae?

All the magnifications listed above 84x will be useless on a 60mm telescope.

Still, such a telescope can be useful if it is well made (particularly a steady tripod/mount) and you have reasonable expectations.

It will show lots of detail on the moon.  You will be able to see the phases of Venus and Mercury (which is pretty much all anyone can see of those from Earth).  You will be able to see Mars as a disk and maybe glimpse an ice cap or a little surface detail under excellent seeing with a favorable apparition.  You will be able to see Saturn's rings and Jupiter's moons and maybe with good conditions a band or two on Jupiter.

It will only show a few of the very brightest nebulae, and only as a faint smudge.


What would the Sun be like if nuclear reactions could not proceed via quantum tunneling? Without quantum tunneling our Sun wouldn't be hot or massive enough to produce the energy it does at the moment. So what would have been the temperature or mass of our Sun without quantum tunneling of protons to maintain the same energy we receive from our Sun?

Short answer: Without tunnelling, stars like the Sun would never reach nuclear fusion temperatures; stars less massive than around $5M_{\odot}$ would become "hydrogen white dwarfs" supported by electron degeneracy pressure. More massive objects would contract to around a tenth of a solar radius and commence nuclear fusion. They would be hotter than "normal" stars of a similar mass, but my best estimate is that they have similar luminosities. Thus it would not be possible to get a stable nuclear burning star with 1 solar luminosity. Stars of 1 solar luminosity could exist, but they would be on cooling tracks, much like brown dwarfs are in the real universe.

A very interesting hypothetical question. What would happen to a star if you "turned off" tunnelling. I think the answer to this is that the pre-main-sequence stage would become significantly longer. The star would continue to contract, releasing gravitational potential energy in the form of radiation and by heating the core of the star. The virial theorem tells us that the central temperature is roughly proportional to $M/R$ (mass/radius). So for a fixed mass, as the star contracts, its core gets hotter.

There are then (at least) two possibilities.

The core becomes hot enough for protons to overcome the Coulomb barrier and begin nuclear fusion. For this to happen, the protons need to get within about a nuclear radius of each other, let's say  $10^{-15}$ m. The potential energy is
$e^2/(4\pi \epsilon_0 r) = 1.44$ MeV or $2.3\times 10^{-13}$ J.

The protons in the core will have a mean kinetic energy of $3kT/2$, but some small fraction will have energies much higher than this according to a Maxwell-Boltzmann distribution. Let's say (and this is a weak point in my calculation that I may need to revisit when I have more time) that fusion will take place when protons with energies of $10 kT$ exceed the Coulomb potential energy barrier. There will be a small numerical uncertainty on this, but because the reaction rate would be highly temperature sensitive it will not be an order of magnitude out. This means that fusion would not begin until the core temperature reached about $1.5 \times 10^{9}$ K.

In the Sun, fusion happens at around $1.5\times 10^7$ K, so the virial theorem result tells us that stars would need to contract by about a factor of 100 for this to happen.

Because the gravity and density of such a star would be much higher than the Sun, hydrostatic equlibrium would demand a very high pressure gradient, but the temperature gradient would be limited by convection, so there would need to be an extremely centrally concentrated core with a fluffy envelope. Working through some simple proportionalities I think that the luminosity would be almost unchanged (see luminosity-mass relation but consider how luminosity depends on radius at a fixed mass), but that means the temperature would have to be hotter by a factor of the square root of the radius contraction factor. However, this could be academic, since we need to consider the second possibility.

(2) As the star shrinks, the electrons become degenerate and contribute degeneracy pressure. This becomes important when the phase space occupied by each electron approaches $h^3$. There is a standard bit of bookwork, which I am not going to repeat here - you can find it something like "The Physics of Stars" by Phillips - which shows that degeneracy sets in when
$$\frac{ 4\pi \mu_e}{3h^3}\left(\frac{6G R\mu m_e}{5}\right)^{3/2} m_u^{5/2} M^{1/2} = 1,$$
where $\mu_e$ is the number of mass units per electron, $\mu$ is the number of mass units per particle, $m_e$ is the electron mass and $m_u$ is an atomic mass unit. If I've done my sums right this means for a hydrogen gas (let's assume) with $\mu_e=1$ and $\mu = 0.5$ that degeneracy sets in when
$$ \left(\frac{R}{R_{\odot}}\right) \simeq 0.18 \left(\frac{M}{M_{\odot}}\right)^{-1/3}$$

In other words, when the star shrinks to the size of $\sim$ Jupiter, its interior will be governed by electron degeneracy pressure, not by perfect gas pressure. The significance of this is that electron degeneracy pressure is only weakly dependent (or independent for a completely degenerate gas) on temperature. This means that the star can cool whilst only decreasing its radius very slightly. The central temperature would never reach the high temperatures required for nuclear burning and the "star" would become a hydrogen white dwarf with a final radius of a few hundredths of a solar radius (or a bit smaller for more massive stars).

The second possibility must be the fate of something the mass of the Sun. However, there is a cross-over point in mass where the first possibility becomes viable. To see this, we note that the radius at which degeneracy sets in depends on $M^{-1/3}$, but the radius the star needs to shrink to in order to begin nuclear burning is proportional to $M$. The cross-over takes place somewhere in the range 5-10 $M_{\odot}$. So stars more massive than this could commence nuclear burning at radii of about a tenth of a solar radius, without their cores being degenerate. An interesting possibility is that at a few solar masses there should be a class of object that contracts sufficiently that nuclear ignition is reached when the core is substantially degenerate. This might lead to a runaway "hydrogen flash", depending on whether the temperature dependence of the reaction rate is extreme enough.

Best question of the year so far. I do hope that someone has run some simulations to test these ideas.

Edit: As a postscript it is of course anomalous to neglect a quantum effect like tunnelling, whilst at the same time relying on degeneracy pressure to support the star! If one were to neglect quantum effects entirely and allow a star like the Sun to collapse, then the end result would surely be a classical black hole.

A further point that would need further consideration is to what extent radiation pressure would offer support in stars that were smaller, but much hotter.


Might the SpaceX Roadster intercept any ongoing surveys? There are star survey telescopes in space, and on Earth. I don't really know how to begin to know where they will all be looking for the next few months. On the off-chance that someone does, or can suggest how to find out in a systematic way, I thought I would ask here.

These is JPL's Horizons predictions for the Roadster spacecraft in its heliocentric orbit. They are based on both the last state vector readings from the spacecraft itself plus several optical sightings my medium sized telescopes.

While the car is fairly small, it's still attached to the white, 2nd stage rocket body, so it continues to reflect a bit of light. You can read more and see some astronomical photos in this answer and read this answer for info.

Well I can't give a mathematically based answer. I can only tell that some man-made objects were recorded by sky surveys. E.g. the Rosetta spacecraft was marked as a newly found asteroid by the Catalina Sky survey although the error was corrected quite quickly.
Rosetta made an earth flyby at this time so it was pretty close.

The SpaceX Tesla will probably not come close to earth for quite a time and as such it probably won't show up in any survey.


What is the large astronomical instrument shown in this 1918 eclipse expedition photo? This nice video musing about (among other things) the US Naval Observatory's $3,500 expedition to Baker City or at least Baker County Oregon to observe the June 18, 1918 solar eclipse, there is a photograph of the expedition showing both normal-sized portable telescopes, and in the background a very long telescope (at least 6 meters, can't see the end) held up by a ladder, with an even longer tent extending backward from the shack sitting at the bottom of the telescope along the ground.

What is this instrument? How does it function optically, and what kind of data does it produce?

below: Screenshot from the video, photo credits available at the end.

This construct was a telescope. Citing this article:

  Owing to the difficulty of erecting a double tower, most other astronomers follow the mechanically simpler plan of laying the camera tube horizontally and allowing the sunlight to be fed into it by means of a plane mirror driven by clockwork to counteract the westward motion of the sun.pg. 255

A Popular Astronomy article also discusses the eclipse observation in detail. They describe the primary telescopes used for photography. The first was a 65-foot telescope (that measurement being the length, not the aperture).

  This instrument was in charge of W. A. Conrad, who was assisted by W. L. Veale. The camera was horizontal and was directed toward the sunrise point. Light from the corona was reflected from a silvered glass mirror 10 inches in diameter attached to a coelostat controlled by clockwork. The lens was 7 1/2 inches in aperture and the plates were double coated special coronal, 17x20 inches, made by the Eastman Kodak Co. The diameter of the sun's image on the plates is about 7 1/4 inches. Five exposures were made of 5, 15, 35, 15, and 5 seconds duration respectively.

They also had a 105-inch telescope which was also laid horizontally and had the light directed into it via a coelostat. While I can't be completely sure which telescope is pictured in that image, given the size, my guess would be the 65-foot telescope.


How is Uranus' north pole defined? The rotation axis of the planet Uranus is tilted by 98° compared to its plane of orbit. This means that the north pole of Uranus is "under" the ecliptic compared to the north pole of other planets.

Why was the value of 98° chosen, rather than 82° ? How was it decided which pole should be north?

From the Report of the IAU/IAG Working Group on Cartographic Coordinates and Rotational Elements of the Planets and Satellites (2000), section 3:

  The rotational pole of a planet or satellite which lies on the north side of the invariable plane will be called north, and northern latitudes will be designated as positive.
  The planetographic longitude of the central meridian, as observed from a direction fixed with respect to an inertial system, will increase with time. The range of longitudes shall extend from 0° to 360°.

  Thus, west longitudes (i.e., longitudes measured positively to the west) will be used when the rotation is prograde and east longitudes (i.e., longitudes measured positively to the east) when the rotation is retrograde.

The "invariable plane" in question is the invariable plane of the Solar System.

By contrast, the usual convention for quoting obliquity is to use the right-hand rule (fingers curl in direction of rotation, thumb points north). This is inconsistent with the definition used to determine the north pole but people seem to prefer having the rotation period always be a positive number.


Occultations of Triton I've been listening to an episode of the "NASA in Silicon Valley" podcast about the SOFIA mission.

One of the techniques they use at SOFIA is stellar occultations to study different objects.

My impression was that the occultations by Pluto happen much more often than by Triton. Is this true and why is that?

Pluto is actually smaller in diameter than Triton, and is also farther away, meaning that Triton covers roughly 1.4x (according to WA) the angle that Pluto does, making occultation that much more probable apriori, ignoring their actual orbits.

In addition to the above, New Horizons recently observed Pluto's atmosphere with far more detail than what we can do from here on Earth, so it makes sense to measure a different target.


Has the permittivity or the permeability of space been measured in a gravitational well? Technically one might say yes since earth is gravitational well. However I am looking for more data points. For example,

in deep space where voyager might be

at the Lagrange points of massive bodies where gravitational fields are in equilibrium

at the bottom of a deep well like Jupiter

The intuition behind this question is the compactification of space that is caused by gravity. Certainly time slows in a gravitational well. And since c is constant, distances are relatively smaller. Please note I am not asking about em waves nor their velocity. Rather the distribution of field lines vs distance from a charge, a moving charge or a magnet.

But you are asking about the speed of light, since the product $\mu_0 \epsilon_0$ determines the speed of light in vacuum, which is fixed.
Before 2019, both the speed of light and $\mu_0$ were defined as fixed numbers in SI units and therefore by definition, $\epsilon_0$ was fixed.
With the revision of SI units, the speed of light is still a defined number but both $\mu_0$ and $\epsilon_0$ are determined by experiment (though their product must still be a defined constant). They are experimentally determined to about 1 part in $10^{10}$ via the fine structure constant $\alpha$. There is no very strong evidence that they or any other physical constants vary in different parts of the universe.
See https://en.wikipedia.org/wiki/Vacuum_permittivity#Redefinition_of_the_SI_units
More fundamentally, it isn't that meaningful to ask whether physical constants with units vary, either in space or time, since they are measured in terms of other things that could potentially vary instead. The only really meaningful question is to ask whether dimensionless combinations of physical constants, such as the fine structure constant $\alpha$, vary.
There have been suggestions that $\alpha$ might vary with time or direction. Since the fine structure constant is
$$\alpha = \frac{e^2}{4\pi \epsilon_0 \hbar c} \ ,$$
then any variation in $\alpha$ could now be interpreted as a variation $\epsilon_0$, since the other constants are fixed in SI units. Or you could rewrite $\epsilon_0$ as $1/\mu_0 c^2$ and interpret it as a variation in the permeability of the vacuum, or some combination of the two.
The results of investigations into high redshift gas clouds along the lines of sight of quasars have proved some of the best probes of this effect. Th emost recent work I have seen (Wilczynska et al. 2020) suggests no variation with time, but there might be a change in direction at the level of 1 part in $10^5$. The present rate of change (if any) must be less than  part in $10^{16}$ per year.


CCD in telescopes: Observation and Astro-photography This is a Celestron CPC 800 XLT telescope. It doesn't seem to have charge-coupled device (CCD) attached. Can a CCD be attached separately to this 8" Schmidt-Cassegrain telescope even if it didn't come along with the package?

For astro-phtography how cheaper and cost-effective would a DSLR camera be than CCD? I think it is better to suit yourself with a CCD than a DSLR camera for quality photography.

Do excuse this novice who is here to learn.

You will typically find that CCD cameras are sold separately to the telescopes. That way the telescope can be used visually or for astrophotography, by either fitting an eyepiece, or a camera. Most telescopes do ship with eyepieces, as they are much less expensive than a CCD and it is sort of expected, but most observers will typically replace this eyepiece with a better one anyway, so to all intents and purposes, when you buy a telescope you are just buying the Optical Tube Assembly and maybe a mount, and you add to that all the accessories you need to accomplish your objectives. Different CCDs will be used for different purposes, for instance planetary observations or deep space objects.

That telescope has a fork mount. Note that this constrains how much equipment you can put on the end of the telescope  without getting in the way of the mount. It also is an Alt-Az mount, which means it is not polar aligned; stars will rotate in the field of view, requiring a field de-rotator in addition to a CCD.

It looks like a nice enough scope for visual observations, but if you are interested in astrophotography, you might want to hold off for a bit, maybe get a scope for visual use first, and maybe make use of well-set-up robotic astrographs that you can hire over the internet, like slooh.com and itelescope.net - get some experience using premium gear that you don't have to worry about setting up or maintaining, and while you do that, work out what gear you will be happy about getting for yourself. The fact you are asking this question about CCDs suggests you aren't yet ready to spend money on astrophotography gear...

Getting set up well for visual stuff can set you back hundreds... Getting set up for astrophotography can set you back many thousands, so you want to be well equipped to make good purchasing decisions. And you can rent a lot of time on some pretty good telescopes for that same $...


Why does a hot cloud need more mass to collapse? I was wondering why does a hot cloud need more mass to collapse than a cold cloud to form a protostar? Is it because there's a higher thermal pressure inside the hotter cloud than it is in a colder one?

It's more do do with having a higher pressure gradient than a higher pressure, though for a cloud of a set size, the two are equivalent.

For a cloud to be in equilibrium requires
$$ \frac{dP}{dr} = - G\frac{M\rho}{R^2},$$
where $dP/dr$ is the pressure gradient and $M$ and $R$ are the mass and radius of the cloud and $\rho$ is its density.

It we just make a rough approximation of a linear pressure gradient and that the pressure on the outside is zero, then
$$ P_c \simeq G\frac{M\rho}{R},$$
where $P_c$ is the central pressure, which is in turn, for a perfect gas $P_c = \rho k_B T_c/\mu$ where $T_c$ is the temperature and $\mu$ is the average mass of a particle in the cloud.

In order to collapse, the RHS has to get bigger than the LHS.  For a given mass, radius and density, then this means the temperature of the cloud must be less than some critical value.  i.e.
$$ G \frac{M\rho}{R} > \frac{\rho k_B Tc}{\mu}$$
$$ T_c < \left(\frac{G\mu}{k_B}\right)\left( \frac{M}{R}\right)$$

Thus a hot cloud won't collapse and a cold one will. But one way to get the hotter cloud to collapse would be to increase $M$ (while keeping $R$ constant).


What's the equation for moon-moon tidal heating? Between a moon and the primary, the equation for tidal heating is:
$$\dot E_\mathit{Tidal} = - Im(k_2) \frac{21}{2} \frac{GM_h^2 R^5 n e^2}{a^6}$$
But how does one calculate the tidal heating between moons?
Simplifying assumptions I'm fine with:

The moons are coplanar, $I_\mathit{affected} = I_\mathit{perturbing} = 0$
Both orbits have no eccentricity, $e_\mathit{affected} = e_\mathit{perturbing} = 0$ (which also implies no heating from the primary)
The affected moon is tidally locked to the primary
The perturbing moon can be treated as a point mass

While I'm not able to come up with any formula, I suspect the following properties hold:

The tidal heating is still proportional to $Im(k_2)$, as this seems to only be an internal property of the moon.
It's still proportional to $R^5$
Since tidal forces are inversely proportional to distance cubed, I think the overall heating is proportional to ${(a_\mathit{affected} - a_\mathit{perturbing})^{-3}}$, due to most heating happening while they are in close proximity.
It's inversely proportional to the relative synodic period of the two moons.

This is for instance the case for the Saturn moons Enceladus and Dione.

This is not the case for Enceladus and Dione. Enceladus's orbit about Saturn has an eccentricity of 0.0047, which while low, is more than enough to result in tidal heating. Dione does play a role in the tidal heating of Enceladus, but that role is secondary. Dione is what keeps Enceladus's eccentricity non-zero via Enceladus's 2:1 mean motion orbital resonance with Dione.


Why can't we see the supermassive black holes supposedly at the center of galaxies? Is gravitational lensing responsible for obscuring (figuratively) the supposed supermassive black holes at the center of our own and other galaxies? By "see" I mean relative to the measuring tools we are using. A radio telescope "sees" in radio waves.

In Hubble photos of galaxies, no black holes appear to be present in galaxy centers, at least not visually. How can this be explained?

The main reason supermassive black holes aren't visible in images of our galaxy is because of all the dust and stuff in between the center of the galaxy and us.  The Milky Way's bulge is ~16000 light-years thick.  There are millions of stars, and huge clouds of dust in between us and the black hole, which block visible light.  Most of our measurements of the center of our galaxy are made in radio wavelengths.

In addition, compared to galaxies, black holes are small.  A supermassive black hole might have a radius of up to 400 AU (for the really large ones).  The nearest spiral galaxy to us, Andromeda, is 778,000 parsecs away.  This corresponds to an angular size of 0.0005 arcseconds.  Even the Hubble Space Telescope only has a resolution of 0.043 arcseconds in the optical.  That means that Hubble can only see things that are 100 times this size.  So we wouldn't even be able to see any black holes in another galaxy.

That being said, we can make observations of black holes using indirect means.  For example, one of the ways we are able to precisely measure the black hole in our galaxy is by watching stars orbit around it.  We take pictures of these stars in the infrared and radio wavelengths, which pass through the dust.  By measuring how fast these stars are moving, we can calculate the mass of the black hole.

In other galaxies, we can observe other effects of black holes, such as active galactic nuclei, and quasars.

So despite being unable to get an optical picture of a black hole, we still know that they're there.


Why can I see sometimes a horizontal half moon instead of a vertical one? Why do I sometimes see a horizontal half moon, instead of the more usual vertical one? I keep getting answers for crescent moons, but this is a half moon with its curve closer to the horizon, like a bowl of cereal.
I would like to know why it only appears sometimes that way.

It occurs because the lit half of the moon points towards the sun, along a great circle in the sky.  The half-moon will be 90 degrees away from the sun in the sky. So by (say) 8pm (at equinox for simplicity) the half-moon will be visible in the South west, but the sun will be below the horizon.  The great circle from the moon to the sun will go "down" relative to the horizon.
The angle that it points down will depend on where you are on Earth. Only near the equator could the moon be completely horizontal,