How fast is a comet moving when it crosses Earth's orbit? Is it about the same as Earth's orbital speed?

Comets don't cross Earth's orbit really. Orbits are one-dimensional objects and their chance of crossing in 3D space is 0. Henceforth, I consider a comet at distance 1AU from the Sun.
What's the maximum speed of a returning comet at 1AU from the Sun? This can be easily worked out from the orbial energy
$$
E = \frac{1}{2}v^2 - \frac{GM_\odot}{r},\qquad\qquad(*)
$$
which is conserved along the orbit ($v$ and $r$ the Heliocentric speed and distance). For a returning comet, $E<0$ and the speed cannot exceed the escape speed (which occurs for $E=0$)
$$
v_{\rm escape}^2 = 2\frac{GM_\odot}{r}.
$$
The speed of the Earth can be worked out from the Virial theorem, according to which the orbital averages of the kinetic and potential energies, $T=\frac{1}{2}v^2$ and $W=-GM_\odot/r$, satisfy
$
2\langle T\rangle + \langle W\rangle=0.
$
For a (near-)circular orbit (such as Earth's), $r$ is constant and we have
$
v^2_{\rm Earth} = GM_\odot/r.
$
Thus, at $r$=1AU
$$
v_{\rm escape} = \sqrt{2} v_{\rm Earth}
$$
as already pointed out by Peter Horvath. Non-returning comets have local speed exceeding the escape speed.
Can a comet near Earth have a speed similar to Earth's orbital speed?. Let's assume a comet with the same speed as Earth at $r$=1AU and work out the consequences. Such a comet must have the same orbital energy as the Earth and, since
$$
E = -\frac{GM_\odot}{2a}\qquad\qquad(**)
$$
with $a$ the orbital semimajor axis, must also have $a=1AU$ and the same orbital period as Earth, i.e. one year. Moreover, the comet's apohelion satisfies
$$
r_{\rm apo}\le 2a = 2{\rm AU}.
$$
Such comets don't exist AFAIK. Most returning comets have much longer periods than 1 year.
What's the typical speed of a returning comet when at distance 1AU from the Sun? To work out this question, let's parameterise the comet's orbit by its period $P=2\pi\sqrt{a^3/GM_\odot}$. From this relation we immediately get
$$
\frac{a_{\rm comet}}{\rm AU} = \left(\frac{P_{\rm comet}}{\rm yr}\right)^{2/3}.
$$
From equations ($*$) and ($**$), we can then find
$$
v_{\rm comet}(r=1{\rm AU}) = \sqrt{2-\left(\frac{P_{\rm comet}}{\rm yr}\right)^{-2/3}}\,v_{\rm Earth}.
$$
In the limit of $P\to\infty$, this recovers our previous result $v_{\rm comet}\to v_{\rm escape}$. For typical period of $\sim$70yr, the speed of the comet is close to this number.
Finally, I shall coment that all this only relates to the magnitude of the orbital velocity (speed), but not to its direction. Comets are typically on highly eccentric orbits and, when at $r$=1AU, move in quite a different direction than Earth, even if their speed is only slightly larger. So the relative speed $|\boldsymbol{v}_{\rm comet}-\boldsymbol{v}_{\rm Earth}|$ of a comet with respect to Earth can be anyting between about 10 and 70 km/s.


Why are there no stars on New Horizons images of Pluto I followed the New Horizons Mission a little, and saw among others this image of Pluto:

I wonder, why you can't see any stars on it. As far as my very basic knowledge in astronomy goes, I think you can even see stars during a "Sun-Moon-Eclipse" here on earth. So besides having here a spectacular "Sun-Pluto-Eclipse", I further more ask myself if that eclipse is even necessary to see other stars, as those pictures are already taken quit far away from sun, and outside an atmosphere.

Is that not enough to see stars? Or does NASA remove the stars from the images, before publishing them? (If so, why? Not to distract?)

Thanks for answers!

The answer by @mso is very good. Another contributing factor is the exposure time of the camera on New Horizons.

The amount of light reflecting from Pluto is greater than that from the background stars. As a result the exposure time is set to achieve a good image for the desired target and detail required.

Over the integration time the amount of light from background stars is not enough to register onto the focal plane and subsequent image.

In order to get good pictures of stars one needs to integrate over long period of time while maintaining accurate pointing.


Hot Jupiters around heavy stars? Hot Jupiters are giant planets that are very close to their host star.
How many Hot Jupiter exoplanets have been found around stars more massive than our Sun?

Has the number of Hot Jupiters discovered as a function of host star mass been plotted? Say $x$ Hot Jupiters around stars that are 1 to 1.5 solar masses ; $y$ Hot Jupiters around stars that are 1.5 to 2 solar masses,...

Yes, there are hot Jupiters around stars more massive than the Sun: in fact the first known hot Jupiter host star 51 Pegasi has a mass of around 1.11 solar masses.

As for a list of them: use the confirmed planets table at the NASA Exoplanet Archive. You just need to add in the relevant criteria you're interested in. The actual criteria for where to draw the dividing line seems to be roughly around Saturn-mass planets, so using Saturn's properties you could do the following:

Filter by mass of planet > 0.3 Jupiters, orbital period < 10 days, mass of star > 1 solar mass. At the time of writing this selects 275 planets.
Filter by radius of planet > 0.8 Jupiters, orbital period < 10 days, mass of star > 1 solar mass. This gets me 286 planets at the time of writing.

There is a handy "Plot Table" button that will allow you to construct a histogram from the selected planets. Here's an example from the planets selected by mass between 1 and 5 solar masses (there are a few outliers at even higher stellar masses, but I'm not sure how reliable those mass estimates are):

Note that the above does not represent the actual distribution of hot Jupiters with stellar mass: there are detection biases at play that make it harder to find planets around more massive stars, which need to be corrected for to estimate the true distribution.

You can also do similar queries at the Extrasolar Planets Encyclopaedia but the user interface isn't quite as friendly.


How does an absolute horizon form before the apparent horizon? If I understand it correctly, the apparent horizon is the boundary between where light directed outwards will move outwards, and light directed outwards will move inwards. Even if light is moving away from the black hole outside of the apparent horizon, it will not escape to infinity if it is within the event horizon.

The event horizon is essentially the apparent horizon in the future, if the apparent horizon continues growing. Theoretically, the two horizons should eventually meet up...again, if I understand it correctly.

My confusion is with how the two form. According to Wikipedia,

  In the simple picture of stellar collapse leading to formation of a black hole, an event horizon forms before an apparent horizon.

My main question is, how can the event horizon form "before" the apparent horizon? I understand that one can slice the Schwarzschild geometry so that there is no apparent horizon at all, but if an apparent horizon exists, I still don't understand how or why it would form after the event horizon.

I guess you are talking about black hole formation here. If we take the Oppenheimer-Snyder model for the spherically symmetric collapse of a star, then an event horizon forms first followed later by an apparent horizon that is at, or interior to, the event horizon.

The event horizon is the surface behind which light rays will never reach an infinite distance from the black hole. The apparent horizon is the surface behind which outwardly directed light rays will move inward.

The two will coincide for a static black hole, but for a forming black hole with a growing mass, it is possible for an outwardly directed photon to be outside the apparent horizon, but inside the event horizon, because the black hole mass is inevitably larger in the future if some of its future mass is still beyond its final Schwarzschild radius. As the event horizon forms at the centre of the star and moves outwards, then it is clear that the apparent horizon, which is always interior to the event horizon, must form later.

In the very simplified Oppenheimer-Snyder model, featuring an initially uniform star comprising of pressureless "dust", the apparent horizon first forms just as the surface of the collapsing star coincides with the event horizon and the apparent horizon is therefore always coincident with the event horizon. In more realistic models the apparent horizon forms a little earlier and then moves outwards to coincide with the event horizon as the surface of the collapsing star moves inside the event horizon.

NB: Note that whilst the apparent horizon can be defined at any point in time, the "event horizon" referred to above is a theoretical ideal, since we cannot preclude that the black hole might accrete some more mass in the future!


Could a star closely orbit a black hole long enough for the star to have lost 0.5B+ years to time dilation? I was wondering how stable a close star-black hole system could plausibly be, and thus how much time a star could plausibly miss out on (from an outside observer's perspective) due to being in an orbit at relativistic speeds around a black hole.

E.g. could there plausibly be a scenario where a star has only aged less than 13 billion years, but is in fact one of the universes's earliest Population II stars, because it has been in a fairly stable, close, rapid orbit around a black hole for the past few billion years*, and thus was subject to time dilation during a few billion years of orbits?

(*Or was orbiting a BH at some point, and escaped due to a chance interaction with a newcomer third body that tossed it out.)

Perhaps this would require a larger black hole, so that the tidal forces would be lower, which would presumably causing less gas stripping for an orbit that would result in the same time dilation? I realize that even well outside the photon sphere of a black hole, there can be decay of orbits due to gravity waves, but evidently there can be a counteracting separating phenomenon well, due to stars being stripped of some mass (e.g. as described in this report of a star orbiting a black hole at a velocity of about %1 of c).

So to put a number on this to make it more answerable, could a star plausibly orbit a black hole long enough and closely enough without spiralling in (or outward due to gas stripping) to make it half a billion years older in origin (according to the CMB rest frame) than it has experienced?

Yes, but not very likely.

The closest orbit that does not require constant expenditure of energy to maintain it is the prograde equatorial ISCO. For a Kerr black hole the time dilation factor on this orbit is $$\frac{dt}{d\tau}\approx \frac{2^{4/3}}{\sqrt{3}(1-a/M)^{1/3}},$$ which at the astrophysically likely Thorne limit $a = 0.998M$ gives a dilation of 11.5470. For a Schwarzschild black hole ($a=0$) it is merely 1.22. (Hamilton 2018, p. 654)

So a 13 billion year black hole could have a star orbiting that had just experienced 13/11.5470=1.12583 billion years!

Now, the gravitational wave emission from orbiting like this would no doubt sink the star. After all, stellar remnants starting at least part of an AU apart are merging today, and the emission scales up fast as one gets closer. Ergosphere processes might happen, but they also reduce $a$ and hence the time dilation.


Is there a prediction when our solar system would fall apart? Within a few billion years our Sun would become a red giant and destroying Mercury and Venus and perhaps even the Earth. But nevertheless our solar system still have some planets like Jupiter and Neptune orbitting around the sun, because his mass wouldn't be very different.

After another while our Sun would turn into a white dwarf. Probably his mass changed a bit, but would it be enough to lose for example Neptune?

If not, is there a prediction at what time, if ever, our 'sun' will lose (some of) his planets for the first time?

The process whereby the Sun will become a red giant and then turn into a white dwarf is neither short nor simple. It's absolutely not the case that it will expand, blow away a chunk of its mass, contract again, and that will be it.

According to Wikipedia (emphasis mine):

  For the Sun, four thermal pulses are predicted before it completely loses its outer envelope and starts to make a planetary nebula. By the end of that phase – lasting approximately 500,000 years – the Sun will only have about half of its current mass.

There's a good lengthy coverage of the last days of the Sun here. It cites a study and then concludes that

  The bottom line is that the orbits are unpredictable and where the
  planets find themselves once the star dies is dependent on when, how
  much and how quickly the sun shed half its mass. This makes a major
  difference in how the solar system will look in the end and how easily
  the planets may be ejected from their orbits.

So the simple answer to your question is: as the Sun loses mass, the orbits of the outer planets will become larger. When they become wide enough, they will escape into interstellar space. If the hypothesized Planet Nine exists, it could help kick out Uranus and Neptune. Jupiter (and Mars, I guess) may or may not remain. Earth will in all likelihood be dragged into the Sun during the red giant phase as the Sun's outer layers expand to fill its orbit.


Compared with similar stars, does the sun itself have properties that make life on Earth possible? We often hear our solar system has properties for life. For example, it is far away from the center of Galaxy. It has Jupiter far away to protect us from asteroids. It has optimum size of territorial planets, plate activity on Earth, and so on.

But besides that, compared with similar stars, does our sun itself also have properties that make life possible? For example, does it have extra stable brightness? Or does it emit less high energy radiation?

One thing that is, if not unique, unusual about the Sun is the size of it.
It belongs to the top 5% brightest stars, meaning that 19 out of 20 stars are smaller than the Sun.

Having a Jupiter size planet is certainly not unique, and plate tectonics we know nothing about for exoplanets.


What is the distance between 2 cities? Both Quito and Kampala lie on the Equator. The longitude of Quito is 82°30'W and that of Kampala is 37°30'E. What is the distance from Quito to Kampala?
A) along the shortest surface path
B) along a direct through the Earth path?
Please help me visualize it with a figure. How to do it?

This is hardly an astronomy question, but I like drawing, so here you are:

Longitudes are measured from the Greenwich meridian, so the angle between Kampala and Quito is
$$\theta_\mathrm{Q} + \theta_\mathrm{K} = 82.5^\circ + 37.5^\circ = 110^\circ.$$
(remember that $0^\circ30' = 0.5^\circ$).
The shortest surface path is along Equator. Since $110^\circ$ is $\frac{110^\circ}{360^\circ} \simeq 0.3$ times the circumference of Earth at Equator, the length of path A (the dashed line) is
$$\mathrm{A}:\,\,d = 0.3 \times 40,075\,\mathrm{km} = 12,245\,\mathrm{km}$$

For path B (the solid line), you need a bit more  trigonometry. The radius of Earth is $R = 40,075\,\mathrm{km}\,/\,2\pi = 6,378\,\mathrm{km}$. The rest will be left as an exercise.


What does it mean that the Sun`s rotation axis is: RA 286.13° , Dekl: 63.87°? first post here.

I found 1 related Question, which has not solved my Problem.

SHORT QUESTIONs VERSION:

Suns axis is 7,25° tilted from the ecliptic. But in which direction?

LONG QUESTION:

Nasa defines the Suns Rotation axis as described in the title:
https://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html

Why is it described with (Ra, Declination)? Is this a (Earth) Aequatorial System placed in the middle of the Sun?

If true, then this should work (What is the angle between the equator of the sun, and the plane of the Earth's orbit). As the best anwer says, the angle between ecliptical pole and sun Rotation axis should be 7.25°. I get 4°, what is going on here? Have I messed up the angle calculation?

Thanks a lot, this is making my Brain burn : D

You can think of the position given in your title as the coordinates of the star that is directly over the Sun's north pole. (Similar to how Polaris is almost directly over the Earth's north pole.) Likewise, the ecliptic plane has a north pole which can be located by right ascension and declination.

7.25° to 7.26° appears to be the correct calculation, based on the following:

North pole of sun: 286.13° RA, 63.87° declination
North pole of ecliptic: 270° RA, 66.5738° declination (per Wikipedia)

$$cos(angle)=sin(dec1)sin(dec2)+cos(dec1)cos(dec2)cos(RA1-RA2)$$
$$cos(angle)=sin(63.87)sin(66.5738)+cos(63.87)cos(66.5738)cos(286.13-270)$$
$$angle=7.26$$


Is the universe like the earth? Is the universe much like the earth, in that if I were to travel in one and only one direction, I'd eventually end up where I started from?

I'm not an astrophysicist but I'll give it a shot. I'm taking this mostly from Hawking's "A brief history of time" and my own personal studies.

First, the universe is not like the earth. As far as I know this analogy only works when one considers the concept of imaginary time. According to Hawking, when you consider the theory of relativity with a imaginary time (which mathematically is pretty much only considering time as a imaginary number, rather than a real one) the equations that describe space are mathematically similar to those that describe a sphere. This is a trick used to try to make a quantum mechanics concept (Feynman's sum over histories) work with relativity.

This, however, does not mean that we can experience space (or the universe) like a sphere, because we are not capable of experiencing imaginary time. We experience only real time. If we were to experience time as it is defined by imaginary time, we would be able to to travel to the future and past as easily as we would travel from latitude -25º to +25º, and we all know that is not the case.

And in this answer, I'm not even taking into account the expansion of the universe, which would complicate things even further for someone trying to go around the universe. Just for the sake of the argument, imagine going around a sphere that is growing more and more. Even if you take the shortest path possible, depending on the rate of expansion the task becomes either impossible of very hard to say the least.

Hope this helps.


What time and where on earth is the latest solar noon? For context: I am visiting Portugal from Australia and after observing that people tend to start their days later, I realised that solar noon also seems to occur at a later time - about 1.20pm at the moment. I don't recall solar noon every occurring at such a late time in Australia. When and where is the latest solar noon on earth?

That should be in western China, since all of China uses Beijing's time zone.

Reference: https://en.wikipedia.org/wiki/Time_in_China

--- Edit below ---

In response to the comment by @adrianmcmenamin: I'll leave this as guesswork since I simply don't know the peculiarities of every timezone there is. Here's a back-of-the-envelope calculation for local noon in western China.

The Sun covers 15 degrees in a period of 1 hour. All of China is on Beijing time (UTC + 8 hours). With the above, it's centred on $15 \cdot 8 = 120^{\circ}$ east. As a consistency check, Beijing is at $116^{\circ}$ east. The westernmost part of China is at about $73^\circ$ east (See here). Local noon there is delayed by $(120^\circ - 73^\circ)/(15^\circ/h) \simeq 3h$.

So, local noon is a little later than 3pm. This is discounting daylight savings time, which is not currently observed in China.


First night on a telescope questions So I recently got into astronomy and I am using my fathers old Meade EQ1-B reflector telescope. This is my first night using it and I have come across a few questions. To begin, the telescope was left alone for several years (the scope is a fairly old model) and may have become damaged over time, yet nothing too severe. We were cleaning it and got to the mirror to find that over the years the "mirror" cover on top of the glass had chipped away all over, giving the result of a patchy look.

We were testing the scope out during the day and the image was fine - the mirror didn't affect it. Now, I am looking upon stars and have noticed a couple of things. When viewing these stars the image sort of moves, not the scope, the image. As if it were an animation, the star isn't just a white blob, it looks like the sun, where the blob is moving - as if it were a liquid. No I am not stupid enough to point the scope at the sun... during night time. The stars just move like a liquid in the center, is this an effect of the mirror? I can't see why it would be as the image was fine during daylight. Is this how all stars work? I haven't had a chance to point it at the moon or any other planets so I can't test it on an object that isn't a star.

Also, as the scope is a reflector it has the "three lines" on the front, however this interferes with the object I am looking at as i can see the lines on the star. Is there any way to stop this, or is it just the result of using a reflector?

Thanks!

Image below:

the "mirror" cover on top of the glass had chipped away all over,
  giving the result of a patchy look.

Optics can take a huge amount of damage (chipping, scratching) before it really starts to affect performance. You'll be surprised at the amount of abuse a telescope can take like that. As long as it's not properly cracked in two, don't lose any sleep over it.

  The stars just move like a liquid in the center

The moving around is normal. It is due to atmospheric turbulence, which happens all the time. Unless you're in a vacuum, the images will always shimmer a little. Sometimes they will shimmer a lot. Other times not so much.

In astronomy we have a word for this - it's called seeing:

http://en.wikipedia.org/wiki/Astronomical_seeing

  Also, as the scope is a reflector it has the "three lines" on the
  front, however this interferes with the object I am looking at as i
  can see the lines on the star.

You are way, WAY, WAY out of focus. The scope might also be completely miscollimated. The star must not appear like a big blob, with the "lines" and the image of the secondary mirror visible in the middle, like in your photo.

The "lines" are called spider vanes - they are parts of the assembly called spider, which holds the secondary mirror in place. The vanes, or the secondary mirror itself, must never be visible in the image.

First, do a rough, purely visual collimation. Do this in a place with plenty of light (inside the house). Take out the eyepiece. Look into the focuser. You'll see the secondary mirror. Reflected in it, you'll see the primary mirror from the bottom of the tube. Reflected again in the primary, you'll see the focuser, with your eye in the middle.

If things are more or less centered, like in the third diagram above, leave it alone. If they are way out of center, like in the first diagram, use the adjustment screws on the mirrors and center everything. Collimation is very important to performance. A miscollimated scope will perform miserably. This is FAR more important than all the scratches on the mirror.

Again, this is purely visual, no-tools collimation. This is all you need for now. You can get a laser collimator later, after all your problems are solved. This is the article where I copied that image from, if you need more info:

http://www.skyandtelescope.com/astronomy-resources/how-to-collimate-your-newtonian-reflector/

Also see this document:

http://www.etnaastros.com/documents/No_Tools_Collimation.pdf

Once collimation is done, you need to find true focus. Go outside at night. Point the scope at a medium-bright star. Polaris, if you can find it, is great because it doesn't move.

Now moving the focuser back and forth, try and determine which way you can make the image of the star appear smaller. Now keep adjusting the focuser until the image of the star is super tiny, then it starts to grow again. Then back off until you make it as tiny as possible. That's when it's in perfect focus.

Now point the scope at a very bright star. If your scope has 3 spider vanes on the secondary mirror, you may see 6 spikes coming out of the star. Like this:

If your scope has 4 vanes, then you should see 4 spikes, like this (it's really 8 spikes, 2 times the number of vanes, but in this case they are superimposed 2 by 2):

The diffraction spikes are easily visible on very bright stars. On medium-size and small stars, they are not easily visible, or not visible at all. This is normal.

Unless the stars are super-tiny dots like in the images above, your scope is not ready for use. Do your best to collimate it and find true focus. Make the stars super-tiny.

Good luck and let me know if you need more advice. It would help a lot if you could find an experienced astronomer nearby to help you fix the scope.


Does anybody know date of publishing of following spirit rover image in NASA apod? I want to find this image in the following linked index.
https://apod.nasa.gov/apod/archivepix.html

It was posted on November 26, 2002, with the caption "Name This Martian Robot."
There was an essay contest to name the two Mars Exploration Rovers to be launched in 2003.
The winner was Sofi Collis, a third grader in Arizona.


Lack of objects between heliopause and Oort cloud? If we take a look at a logarithmic scale of the Solar System, there appears to be a large gap between the heliopause and the Oort cloud:

Why is that? There are some objects in the gap, but why are most of the Oort cloud objects in one area? Shouldn't it also fill the large gap in the picture just as much?

There may be Sednoids there.

Sednoids are a hypothetical class of "inner Oort Cloud objects" named after their prototype, Sedna. Sedna's aphelion is ~936 AU, bringing it close to the inner boundary of the Oort Cloud. Sednoids may have aphelions ranging from about 100 AU to 1,000 AU.

The problem is, only two Sednoids have beet detected to date, 90377 Sedna and 2012 VP113. Brown et al. 2004) suggested that ~500 may be detectable; surveys simply haven't tracked objects in that area.

Why are Sendnoids where they are? Three ideas have but put forth:

A planet at ~70 AU scattered these objects into elliptical orbits.
A close pass by a nearby star.
Interactions with other stars in the Sun's original cluster.

These objects would fill in the space between the Kuiper Belt/scattered disc and the Oort Cloud itself.


How can I work out longitude and latitude through my telescope? We purchased a Celestron Astromaster 90EQ Refractor Telescope and a Celestron Powerseeker Accessory Kit about four years ago but have not used it much for many reasons. As such my husband and I are complete beginners and was hoping for some advice please?
My Dad passed away during the lockdown and we have had a star named after him.
It's a "UK" based star and we will be sent longitude and latitude co-ordinates enabling us to search for it.
On Christmas day we are hoping to use the telescope, with our children, to point out my Dad/Grandad in the sky. However, my husband has now said he thinks you have to buy an accessory or something to do this and the telescope and accessory pack we have does not have the necessary equipment ?.
Can anyone advise how you can do this please and, if we need to purchase an accessory, recommendations?
Thank you very much for any help in advance.
Sue Davis

I used to have this exact telescope. It took a lot of practise, but I was eventually able to reliably find most targets I was after. I used this scope for about 2 years learning my way around the night sky before moving on to another scope and mount. You should practise finding bright targets before moving on to dimmer ones, just to get used to the scope. However, you can use the setting circles on the mount to help you too. The setting circles are the 2 silver disks with all the numbers on the head of the mount. The Declination one is just underneath where the telescope is fitted (upper red circle in the picture). The Right Ascension is on the other rotational point of the mount head (lower red circle in the picture)

First, you need to reliably polar align the scope, which will make nailing down other objects easier. This may take some time so for a "roundabout" method, you can just point the mount towards polaris and it'll be good enough to keep objects centered when adjusting the RA knob, but I'll still go through the "drift" method for a more reliable alignment.
Drift aligning on a manual EQ mount can take a while, but it starts with a rough polar alignment by eye. Align the mount with Polaris as best you can. Now, move the telescope and find a bright star to the south. Center the star in your eyepiece then get the highest power eyepiece you have (will probably be the 9mm one in your accessory kit) and move the star so it is sitting nearly top center of the eyepiece. Wait 2 minutes or so, then move the RA knob slowly to bring the star back into view. If the star has drifted up, then your alignment is too far west. Give it a small adjustment to the east. If it drifts down, you are too far east. Adjust it westwards. Repeat until you have minimal drift.
Next, find a bright star in the east. Use the same method of centering a star then moving it to the top of the eyepiece. If this start drifts up, your latitude setting is too high, so move it down slightly. If it drifts down, your latitude settign is too low, so lift it up slightly. Repeat till minimal drift and you have a pretty good polar alignment!
Now to actually find a target!
Find yourself a starting star, which is the brightest star in the nearby area of your eventual target. Center it in the eyepiece and look up the co-ordinates on Google or any astronomy app you may have. Adjust the setting circles so that the 'pointers' point to the corresponding values in the RA and DEC axis. Do another quick check that the star is centered in the eyepiece and the setting circles are correct. Now find the co-ordinates of your target. Loosen the declination adjustment on your mount and move the telescope s that the pointer is pointing at the declination of your target. tighten the clutch. Now loosen the RA and move the scope to the correct RA co-ordinate. Tighten the clutch once there to secure the scope. Using a low power eyepiece (your 20-25mm one that came with the scope will likely be the lowest power), have a look through the scope and your target should be somewhere in the FOV.
It should be noted that this is isn't a 100% reliable method, and the setting circles do have markings very close together so accuracy can be difficult, which is why it is best to use a low power eyepiece to find your target, but it should get you very close. It may be best to practise by starting on a bright star and using this method to find another bright star. Once you get the method, you'll start getting more accurate and speedy with it.
I hope this helps.


Meteor traveling through atmosphere without hitting Earth Is it possible (however unlikely) to have a meteorite miss Earth so narrowly and at such flat angle that it would enter atmosphere, travel few (dozens, hundreds, thousands?) kilometers getting as close as few kilometers to the ground without hitting anything and then leave atmosphere without any significant, leaving millions of humans staring agape at the sky?

Or does gravity and air friction prevent this by either making such flat trajectory impossible or by slowing down meteorite enough to make it fall to the ground?

yes it is possible,they are called Earth grazers or Earth-grazing fireball.
they are not rare but only few incidents are recorded.

for more details read this Wikipedia page https://en.wikipedia.org/wiki/Earth-grazing_fireball


Is Hawking radiation detectable? Could Hawking radiation be detected with instruments available today? If not, what sort of instruments would be required to observe it? Would there be any way to confirm through observations the existence of Hawking radiation?

I'm primarily asking about techniques that we could reasonably implement with current technology or something that could be implemented in the near future, but if you need to get close to a black hole to be able to observe the radiation and couldn't do it from a spacecraft in Earth or solar orbit, then that would be the answer.

No. Problems:

Hawking radiation would have very small temperature of $10^{-6} M_{sun}/M_{BH}$ Kelvins. Primordial black holes below 1kg already did evaporate, so there would be no sources from small black holes. The only hope is artificial black holes created by CERN.
Falling matter inside black hole would expirience enormous amount of acceleration leading to emittance of photons (particularly X-rays). So you would first are required to cut out this noise.
Black holes are surrounded by ergosphere consisting of hot plasma, which effectively blocks all photons below some critical temperature. For example, we have our radiowaves blocked in the same way by ionosphere of Earth.
(Reasons above were not enough?) There is other quantum-mechanical mechanisms of creating particle-antiparticle pairs which potentially lead to heavy emittance of antiparticles, mostly positrons from black holes. So you would somehow need to shield from positrons, and the photons coming after $e+e^{+}\rightarrow \gamma + \gamma$ kind of reactions.

You would be required somehow prepare black hole for investigation by removal surrounding physical layers from ergosphere, stopping its rotation (really impossible?). I don't know of any "surgical" instruments available today for such task ("surgical" in a sense that we will need to modify our black hole).

Our measuring instruments are totally capable, though. We also know where black holes are located. Tiny problem of communication still persists. But I think scientists would be perfectly fine to wait few millions of years for station to return. Also there is problem from the field of mechanics. It is enormously hard to cross ways with gravitationally strong object, because your trajectory (hyperbolic type) would intersect with a lot of other trajectories (of elliptic type). But I did not list it, counting on humankind getting smarter with this kind of things in a few centuries.


Do the E and B modes of the CMB polarization have anything to do with electric and magnetic fields? As explained in this article, as the CMB is created from a vaiable (matter?) density $\kappa$ its gradient $\nabla \kappa = \vec{u}$ can be separated into a gradient and curl part, called the $E$ and $B$ modes

$$
\nabla^2 \kappa^E = \nabla \cdot \vec{u}
$$

and

$$
\nabla^2 \kappa^B = \nabla \times \vec{u}
$$

Thinking about electromagnetism, these definitions look slightly (but not exactly) familiar, if the vector $\vec{u}$ would take the role of the electromagnetic potential $\vec{A}$.

Is the similarity to the terminology of electromagnetism just a coincidence, or do these $E$ and $B$ modes have indeed something to do with electric and magnetic fields ?

In this context, the $\kappa$ you are referring to is called the dimensionless matter density field. It is gravitational lensing jargon, and is usually just referred to as the 'convergence' field.

What you have written there (that a field can be split up into divergence-less and curl-less components) is generally speaking true for any field which can be expressed as the curl of another field $A$. @chris is correct in that they refer not to the electromagnetism you are probably used to hearing these types of things discussed in, but refer to the polarization states of the cosmic microwave background due to quantum fluctuations in the early universe. They are called 'E' and 'B' only because they are reminiscent of electric and magnetic field lines.

The big find was the measurement of B-modes. As a reminder, here is what these E- and B-mode patterns in the CMB would look like:

The big thing here is that E-modes are symmetric upon reflection, whereas B-modes are clearly not ($B>0$ maps into $B<0$). In order to produce B-mode polarization, you need two things:

1) Thompson scattering of photons by free electrons (present in the surface of last scattering as well as reionization).

2) A temperature quadrupole.

The exciting thing about the second item is that there are only a few things which can produce just such a quadrupole moment, and one of them is gravity waves! Gravitational waves physically warp the coordinates of space-time as it travels through space, which of course have existing particles, causing these types of distortions:

This is the source of the temperature quadrupole moments. Sides which are squeezed are hotter and sides which are stretched are cooler.

Gravitational lensing can also produce small amounts of B-mode like features in the cmb polarization, but I believe they occur on much smaller scales and can therefore be ruled out as the cause of the BICEP2 findings. Lensing tends to stretch things out perpendicular to the radial vector (in the $\hat{\theta}$ direction):

The reason why this matters is because lensing provides a mechanism for turning E-mode polarization (take the $E<0$ mode for example) into a B-mode type polarization. Most plots of the BICEP2 results show the effect of lensing at much higher multipole moment (i.e. - smaller physical scales), so I believe they have a good way of differentiating between the two types of secondary cmb anisotropies.


When was the last volcanic eruption on Mars? There are several volcanoes on Mars, including Olympus Mons, the largest volcano in the solar system, the volcanoes on the Tharsis bulge,... But they are all thought to be inactive now.
When did the last volcano erupt on Mars?

Berman and Hartman (2002) dated some lava flows of the Athabasca–Marte Valles system at < 20 Ma. Later, Vaucher et al. (2009) dated lava flows of the same region, finding ages even younger in some cases (around 2.5 Ma). In both studies the age estimates are derived from surface morphology (crater counts).
In Volcanoes: Global Perspectives, Lockwood & Hazlett go even further (emphasis mine):

Unlike Earth’s Moon, volcanic activity may be far from over on Mars. Detailed photographic studies by NASA’s Global Surveyor starting in 1997, and the Mars Express High Resolution Stereo Camera, which began orbiting Mars in December of 2003, provide evidence of effusive volcanism perhaps as young as 2–2.5 million years, with flows mantling the flanks of Olympus Mons erupted mostly within the past 200 million years. It cannot be assumed that Mars is now internally cold and volcanically dead, and in fact, volatile efflux may still be in progress.


Starting a fire in a cold planet that was full of flammable gas What would happen to a cold planet that has a large amount of flammable gas, like Neptune, if we throw some fire into it? Will it burn, or would the flame be extinguished due to the cold?

The outer parts of Neptune are mostly hydrogen and helium. There are small amounts of other gases such as methane, ammonia and water vapour. However, there is no oxygen at all.

If you took some of Neptune's outer layer back to earth and mixed it with our air, it could burn. Even very cold hydrogen can burn (it soon heats up!)  This couldn't happen on Neptune, because a fire needs both fuel and oxygen to burn.

It is very unlikely that any planet would have large amounts of both fuel and oxygen in its atmosphere. Oxygen is very reactive and will react with any flammable gases to produce (mostly) water and carbon dioxide. Oxygen is not stable in a hydrogen atmosphere over a period of millions of years.  If we find oxygen in a planet's atmosphere, we can be fairly sure that something on the planet is making it.


Do other planets with a magnetic field also have two Van Allen belts? On Jupiter and Saturn, aurora is seen around the planet's poles, so that implies that there must be a kind of magnetic field around those planets. Perhaps even the moon Europa could have a magnetic field created by salty water (Does it?).

On Earth we have two radiation belts, an inner and outer belt. One contains electrons and the other mostly protons. Do other planets also have these two radiation belts?

Yes, Jupiter and Saturn have their own radiation belts.  Jupiter's is extremely powerful, which is why all of Juno's sensitive electronics had to be radiation hardened and shielded inside a vault.  It's also why Juno is going to be on a polar orbit doing flybys of Jupiter, instead of orbiting around it closely.

These magnetic bands are indeed separated into inner and outer portions.  From Wikipedia:

  Jupiter's magnetosphere is traditionally divided into three parts: the inner, middle and outer magnetosphere. The inner magnetosphere is located at distances closer than 10 RJ from the planet. The magnetic field within it remains approximately dipole, because contributions from the currents flowing in the magnetospheric equatorial plasma sheet are small. In the middle (between 10 and 40 RJ) and outer (further than 40 RJ) magnetospheres, the magnetic field is not a dipole, and is seriously disturbed by its interaction with the plasma sheet (see magnetodisk below).
  Jupiter's inner magnetic belt was mapped by Galileo, showing it contains high-energy electrons.  Eruptions from Io result in the middle portion of the magnetosphere containing huge amounts of ions of various elements.

Saturn's magnetosphere is likewise divided into separate parts.  However, there are 4 regions to Saturn's magnetic field, and they are separated by the kinds of particles they contain.  These particles result from interactions with Saturn's various moons, as well as its ring system.

The Galileo probe discovered that Europa does produce a weak magnetic field, as a result of its subsurface ocean.  This ocean has an induced current created from Europa moving through Jupiter's magnetosphere, which in turn creates a magnetic field.  This magnetic field is actually one of the things which caused many scientists to theorize a subsurface ocean.

However, these belts are not "Van Allen" belts.  That name is reserved strictly for Earth's radiation belts.


is there any mathematical formula, paper or anything that shows the probability of a meteor hitting a particular spot of moon's surface? According to this article http://news.nationalgeographic.com/2016/02/160209-meteorite-death-india-probability-odds/ a science professor at Tulane University made a paper to calculate the probability of a person being hit by a meteor on Earth . And also an astronomer made a calculation (not mentioning the procedure) with a similar figure. My question is, since the moon has no atmosphere and more meteors reach the surface, is there any mathematical formula, paper, etc. to calculate how likely a spot in moon's surface it's likely to be hit by a meteor?

There is this source that claims 2800 kg of meteor material lands on the moon everyday.  They break down that amount to determine an average area per strike. There have been other studies like this one and this report.  I sure can't vouch for the accuracy of this first report but it's from a reputable (NASA) source.


If an object with mass were to somehow go the speed of light, would it destroy the whole universe? Would an object with mass traveling the speed of light destroy the whole universe because it would have infinite energy / mass?  According to wikipedia and other sources, it takes infinite energy to go the speed of light, which should be impossible.  However, for the sake of the question, let's assume it were somehow achieved e.g. potentially as described here.

My premise behind it possibly destroying the whole universe is that an object with infinite energy should also have infinite mass ($e=mc^2$, and infinity / $c^2$ is still infinity).  Infinite mass should have infinite gravity, which may well destroy the entire universe (although, would this destruction travel at the speed of light, so it would "slowly" destroy the universe vs. instantly?).

One thought is that it could just form a black hole instead, but as I understand it, a black hole is an infinitely small (singularity) of infinite density, with a "net mass" equal to the net mass that it has ingested.

Would an object with mass traveling the speed of light destroy the whole universe because it would have infinite energy / mass?

If we understand the question as a limiting process, which is the only way it makes any sense, the answer is no. For illustrative simplicity, take a spherically symmetric isolated body, so that its exterior gravitational field is the Schwarzschild spacetime. Now boost this to ultrarelativistic speeds. In the speed of light limit, the result is a linearly polarized axisymmetric gravitational pp-wave, the Aichelburg-Sexl ultraboost.

Its effect on test particles it passes is an impulse: it instantaneously bends the worldlines, but does not destroy anything. You can think of this as infinite force acting for an infinitesimally small time, making the overall effect finite and nondestructive, somewhat analogously to a particle encountering a Dirac-delta potential in Newtonian physics.

This is similar to the ultrarelativistic limit of a moving electric charge. The electric field lines are Lorentz-contracted along the direction of travel, squeezing them together (image credit: wikipedia):

In the ultrarelativistic limit, the electric field in the transeverse direction becomes infinitely strong, but it is also Lorentz-contracted to be infinitely thin. Without ignoring the magnetic field, the result is an electromagnetic plane wave with a Dirac delta profile. The gravitational case is analogous, though quantitatively different. Some details on this can be found in arXiv:gr-qc/0110032.

  My premise behind it possibly destroying the whole universe is that an object with infinite energy should also have infinite mass (E=mc^2, and infinity / c^2 is still infinity).

The concept of 'relativistic mass' is largely deprecated in modern physics. Mass is more properly related to energy and momentum via $(mc^2)^2 = E^2 - (pc)^2$.

  One thought is that it could just form a black hole instead, ...

No, that's completely wrong. It doesn't become a black hole for the same reason it doesn't destroy the universe: the gravitating body moving relativistically past a stationary observer is physically equivalent to a relativistic observer moving past a stationary gravitating body. But the observer moving relativistically obviously does nothing to the gravitating body!


What are infall signatures in stars? I came across several papers (e.g. Looking for outflow and infall signatures in high mass star forming regions P. D. Klaassen, L. Testi and H. Beuther, 2011) talking about infall signatures in star formation. Can you please explain the physical meaning of it?

The paper you linked is specifically trying to understand massive $(M\gtrsim 8\:\mathrm{M_\odot})$ star formation. It is still an open question as to exactly how very massive stars form. Do they form all at once in a giant accretion event ("top-down" formation) or do they form by having a bunch of low mass stars merge ("bottom-up" formation) or by some other process altogether? The authors of the paper state that it is particularly tricky to study high mass star formation.

  Evolution in low mass systems is often characterized by changes in the spectral energy distributions (SEDs). This is not as easily done in high mass systems (Molinari et al. 2008)1, which is why studying the gas dynamics is very important.

Basically, they can't easily figure out what's going on by looking at the SED of the high mass star forming region, so instead, they're going to look at how the gas is flowing and moving during the formation. Klassen et al. break their observations into observing "infall" and "outflow" (via two different molecular transitions). What they mean by this is that one molecular transition (that of the $J=4-3$ transition in $\mathrm{HCO^+}$ and $\mathrm{H^{13}CO^+}$ molecules) will show evidence of gas flowing into the core, i.e. infall, while another molecular transition (that of the $J=8-7$ transition in the $\mathrm{SiO}$ molecule) will show evidence of gas flowing out of the core, i.e., outflow.

During star formation, you tend to get both types of flows. Infall will occur because the gas is simply accreting onto the core due to gravity. Outflow will occur because the accreting matter and central core get very hot, causing an energy release outwards.

You can see precisely how they track infall flows in Section 3.2. They have a paragraph discussing precisely how they determine that the gas is falling into the core from the observations of the $\mathrm{HCO^+}$ and $\mathrm{H^{13}CO^+}$ molecules.

  Double peaked or asymmetric H$\mathrm{HCO^+}$ line profiles in which the brightest emission is blueward of the source rest velocity can be interpreted as due to infall if the $\mathrm{H^{13}CO^+}$ is single peaked at the rest velocity. If the optically thin $\mathrm{H^{13}CO^+}$ also has a double peaked profile, the double peak in both lines is likely due to multiple components along the line of sight. Churchwell et al. (2010)2 describe the different mechanisms that could produce a double peak in an $\mathrm{HCO^+}$ line profile towards massive star forming regions, and conclude that self absorption best fits observed line profiles. The blue or red skew of the absorption profile can then be used to distinguish infall (blue) or outflow (red).

I won't go into detail to precisely explain the spectroscopic interpretation of the above quote, but suffice it to say, that is their method of detecting matter infalling towards the core of the high mass star.

1 Molinari, S., Pezzuto, S., Cesaroni, R., et al. 2008, Astronomy and Astrophysics, 481, 345

2 Churchwell,  E., Sievers,  A., & Thum, C. 2010, Astronomy and Astrophysics, 513, 9


Assuming 2007 OR10 is in tidal lock with its satellite, what is its mass? Its rotation period is 45 hours. Its satellite is at the distance of 15000 km. Given this data, and assuming the satellite's orbital period the same as the main body's rotation period, 45 hours, what is the mass of the main body?

We can use Kepler's Third Law, the Law of Periods, to solve this problem. The law can be stated mathematically as:

$T^2 = \frac{4\pi^2}{G(M+m)} a^3$

Where

$T$ is the orbital period,

$G$ is the gravitational constant,

$M$ is the mass of the primary,

$m$ is the mass of the secondary, and

$a$ is the semimajor axis.

Solving for $(M + m)$ with the given values returns a system mass of approximately $8 \times 10^{22}$ $\mathrm{kg}$.

Other data are necessary to isolate the mass of the satellite from the primary.


Velocity of ringularity How fast is a ringularity1 spinning? Spinning black holes are created from spinning stars.
The angular momentum of the 'parent' star must be conserved so spinning black holes are spinning much much faster than their 'parent' star.
Since black holes have generally less mass than their parent star and mass is found only in a tiny amount of space then the ringularity should be spinning incredibly fast.
The ringularity is a ring which contains all the mass of the black hole and it is in the center and it is almost infinitely tiny so it must be spinning almost infinitely fast.
Can it spin faster than c? Is there any way to know if SR can be violated inside such extremely dense objects?
1Ring singularity

$$
    K_\text{min}=2\pi\sqrt{a^2+3(ma^2)^{2/3}}
\tag{1}
$$
where the quantities $m$ and $a$, both of which have units of length, are defined by
$$
    m=\frac{GM}{c^2}
\hskip2cm
    a=\frac{J}{Mc}.
\tag{2}
$$
The maximum spin is a=m (an extremal Kerr black hole), the borderline case between a black hole and a naked singularity.
More details from Chiral Anomaly here:
https://physics.stackexchange.com/a/469282/


Helium in terrestrial planet atmospheres? When you look at effective temperature ranges and escape velocities of planets, there is a "population" where hydrogen escapes, but helium doesn't. Is it feasible that there may be planets less that about 10 Earth masses (not gas giants) with a predominantly helium atmosphere??

Another way of asking this question would be:
Do any models show terrestrial planets retaining a primordial atmosphere or is expected that most all will be secondary? I am guessing that very little helium is "outgassed", whereas hydrogen can be added to an atmosphere by photodissociation and other processes. I know that early star UV emissions, collisions, etc. will probably strip a "close in" planet's original atmospheres.

Do you have any reputable estimates as to how much helium has been out
  gassed over the last 4 billion years

Reputable estimates - no.  Bad guesses, maybe.

Short answer:   only trace amounts.  It's impossible to have enough rare radioactive material to produce a real helium atmosphere.  Even if the planet retains all it's helium from radioactive decay it should be measurable in parts per million, not a real atmosphere.

Long answer below:

The number of alpha particles that a radioactive element creates is pretty straight forward, just subtract the Baryons in the final state from the initial state.   Beta decay changes the type of hadron, but not the number, so, from Uranium 238 into Lead 206 - loss of 32 baryons, and all of them through Alpha decay - Alpha particles become helium atoms, so each U-238 decays into 8 helium.

Thorium 232 into Lead 208 - 6 helium.

Uranium 235 into Lead 207 - 7 helium.

Potassium 40, the other common radioactive element only undergoes beta decay into CA40 (89.3%) and AR40 (10.7%), so it's irrelevant in the helium calculation.  There are smaller amounts of other elements that emit alpha particles, like PT-190, but they're rare enough to be ignored.   The bigger unknown is how many elements with shorter half-life were common when the Earth was young but are no longer present.   I have no idea how to calculate that so that will be a bit of an unknown.  One possible approach would be to say that the forming planet would be sufficiently hot to lose much of it's helium, but that's a bit of a cheat.   I think, early on, predictions are rough due to unknown quantities of shorter-half life elements.

U-238 is easiest to calculate with a half-life of about the age of the Earth, so about half of Earth's U-238 has completed its decay change and produced 8 helium per U-238.

Thorium-232, with a half life of about 14 billion years, so only about 20% of the initial thorium-232 has decayed over the age of the Earth.

and U-235, with a half-life of about 700 million years, means that almost 99% of it has decayed over the age of the Earth.  Current estimates say that about 0.72% of Uranium is U-235

I touched on this above, but the big unknown for this scenario is in the range of 100 million year half-life.   That is, elements that are long-lasting enough to survive the Earth's formation but wouldn't remain in high concentrations today.   That's in this range, bottom of the 10^12 seconds, top of the 10^15 seconds.

Curium-247 has a half-life of about 15 million years and it decays into lead-207 releasing 10 alpha particles in the process.   If planet formation happens relatively quickly, that could supply some helium to a planet, but I'm going to ignore it as having slightly too short a half-life.

Uranium-236 has a half life of about 23 million years and it decays into Thorium, emitting just 1 alpha particle.  Presumably much of it would become thorium by the time of planet formation, so I'm going to ignore it as well.

Plutonium-244, mostly synthetic, but it is the rarest primordial element on Earth.   From the Wikipedia article:

  Accurate measurements, beginning in the early 1970s, have detected
  primordial plutonium-244, making it the second-shortest-lived
  primordial nuclide after 146Sm. The amount of 244Pu in the pre-Solar
  nebula (4.57×109 years ago) was estimated as 0.008 of amount of
  238U.

If estimates put it at a bit less than 1% as common as U-238, I'm going to ignore as well, as 1% won't make a big difference.

Finally, Samarium-146 is worth considering with a half-life of about 100 million years (or 68 million depending on which source you use).   It decays into Neodymium-142 releasing a single alpha particle.   Samarium 147 and 148 also undergo alpha-decay but very slowly, half-life of 100 billion and 7 quadrillion respectively.

So these are the 4 primary helium "outgasing" isotopes relevant after planet formation.    U-238 (8 helium), U-235 (7), Th-232 (6) and SM-146 (1).    And as for how much, you have to ask how long the planet has been around.

SM-146 decays fairly quickly, half life 100 million years.
U-235 - 700 million years
U-238 - 4.5 billion years
T-232 - 14 billion years.

If we take a planet about the age of Earth

SM-146 - essentially all of it
U-235 - 99%
U-238 - 50%
T-232 - 20%

Now the trick is, how much of these elements is a terrestrial planet likely to have at formation.

If we go by radioactive decay and radiogenic heat estimates, TH-232 and U-238 combine to generate 20 trillion watts (first article).  That' 6 x 10^20 joules per year or about 3.75 x 10^33 mEV per year, by radioactive decay of those 2 elements.

The Thorium decay chain emits about 42.6 MeV per decay for 7 alpha particles.  (about 7 MeV per helium) and the U-238 decay chain, about 51.7 MeV for 8 alpha particles.  (ballpark 6.5 MeV).   That includes Neutrinos though and that energy needs to be factored out.   I couldn't find precise numbers but about 10 MeV (or 20%) is lost into Neutrinos (technically anti-neutrinos) by this website.   So, very ballpark, lets say 5 MeV heat creates one alpha particle.

current rates of decay of Thorium & U-238 should produce about 7.5 x 10^32 alpha particles, which corresponds to about 1.2-1.3 billion moles of Helium or about 5 million KG, between 5 and 6 tons of helium produced in the Earth's crust and mantle every year - based on radioactive measurements which are probably pretty accurate.

Thorium is currently about (ballpark) 4 times as abundant as U-238 but because it decays 3 times more slowly the helium production and heat generation in the Earth's core is similar (this chart suggests thorium's heat output is higher, but they are close).

The helium production from Thorium has been pretty consistent, as 80% of the Thorium at the formation of the planet is still here.
The production from U-238 was about twice as high when the Earth formed.
The production from U-235 is currently negligible compared to the other two, but was much higher when the Earth was young.   working back from U-235 being about 0.7% of the total Uranium and it having gone through between 6 and 7 half-lifes since the formation of the Earth, (and accounting for U-238 being twice as abundant when the Earth formed), that means between 32 and 64 (lets say 50) x 0.7, so there was some 30-40% as much U-235 as U-238 when the Earth formed.

Bad math again:

Thorium, 2.5-3 tons per year now, 25% more than that when Earth formed.  lets say 3 tons per year, average

U-238, 2.5 tons per year now, twice that when the earth formed.   lets ballpark average that out to 3.5-4 tons per year on average.

I'm not going to average out U-235 because it starts out too high and decays too quickly but if there was about 30-40% as much U-235 as U-238 when the Earth formed I can just multiply U-238's numbers by 35% and by 7/8, working out to about 1 ton per year average.

I'm doing per year/on average just for quick bad math, so 3 + 3-4 + 1 = about 7.5 tons per year x 4.5 billion years - quick estimate for how much Helium was produced by those 3 elements - very bad estimate, 33 billion tons.   Most of this Helium is produced and would be trapped in the Earth's interior until plate tectonics allows it to escape.  The rate of escape is above my pay-grade.

Samarium-146 is the hardest one to estimate.   It's essentially all decayed into Neodymium-142 but both Samarium (it has several stable isotopes) and Neodymium are abundant, and this chain produces a single helium.   It likely doesn't make a huge difference in the total.   Table of elemental abundance in Earth's crust.

Atmospherically speaking, 33 billion tons is nothing.   Even if we adjust to to 40 or 60 or 300 billion tons, that's still essentially too little to be an atmosphere.  Helium is currently about 5.2 ppm in the atmosphere and giving the Earth's atmosphere a mass of about 5,600 trillion tons, 5.2% by molecule, adjusted for the mass ratio of helium (4) to Nitrogen/Oxygen (29), there's about 4 billion tons of Helium in the Earth's atmosphere today.   If we were to say that 400 billion tons of helium was produced on Earth by radioactive decay (about 10 times my estimate), that's still a trace amount, just 100 times the current amount or 520 PPM, which is less than the Argon (currently 9,300 PPM).   Argon-40 which comes from the radioactive decay of Potassium-40 (about 10% of the time), is more common because Potassium-40 is much more common than the Uranium, Thorium and Samarium-146 that decay alpha-particles.

That's a very long answer that basically says there's not enough Alpha-particle emitting elements that last long enough to survive planetary formation, that produce anything more than a trace amount of helium.

Earth also retains some terrestrial helium, about 7% of Earth's helium is terrestrial, 93% formed by radioactive decay.   Most of Earth's terrestrial helium is in it's mantle, very little in it's crust.

Some interesting and loosely related reading.  Yellowstone is able to release more helium than the radioactive material produce because it's crust is unique, very old and it's trapped helium for over 2 billion years, only recently started to release it.

As to your original question.  As I understand it, a planet needs to be very large to acquire significant amounts of helium.     By particle, the universe is about 92% hydrogen and 8% helium (or 75%-25% by mass).

Jupiter's outer atmosphere is slightly helium richer than that, about 10% helium.

Uranus' uppermost atmosphere is about 15% helium, but only it's outer edge.   Uranus, unlike Jupiter and Saturn, isn't primarily a hydrogen-helium planet, it's more of a water/ice planet, sometimes called an ice giant.

It seems that planets need to be very massive to obtain significant amounts of the most common elements, hydrogen and helium, then, as the Wikipedia article implies, they need to get hot to lose their hydrogen, so helium atmosphere planets probably need to be quite large, perhaps more massive than Neptune.   It might be quite a it more difficult for a typical terrestrial planet to do that, unless if it was very massive and a terrestrial planet that massive might resemble a gas giant anyway due to a very thick atmosphere.

I have a tough time thinking of any other way a helium atmosphere is likely to happen, unless you got the planet below 20 degrees K and all the other gases turned to liquid or ice, but I suspect that's not the answer you're looking for.  A planet probably needs to be quite large to retain significant amounts of helium I would think. outside of a very odd circumstance like a helium white dwarf colliding with another large object inside a stellar nursery, even then, helium gas doesn't clump like ices and rocky material, but maybe such a scenario could put enough helium into the planetary formation.  (bit of a crazy what-if), and I'm not sure I buy it.

I tried to reason this one out so, corrections are welcome and encouraged.   I'll try to clean up any grammar a little later.


Question about the Sun's Motion My name is Chris, I am a junior and I'm taking an Astronomy class in my high school.
For a homework assignment, I have to explain how the Sun would move across the sky from September 1st through September 30th (assuming that the weather allows viewing)

For example:
In this picture the

sun is facing west.
If I were to take a picture of this same area at the same time every day for the whole month,
How would the sun change? Would it go east?
Would it go lower in the sky?

I, being a new student to Astronomy, honestly have no clue.

Thank you and if you need any more info please leave a comment.

Over the year, the sun follows a path known as an annalemma. This is a figure eight shape and is caused by the combination of the Earth's rotation around the Sun and its tilt relative to the plane of the equator.

The analemma can be observed by pointing a camera at the sun, and at the same time each month, take a picture and over lay the images:

Taking a picture every day for a month (in September) would mean that you'd see it lower in the sky, curving slightly as it followed the figure of eight shape of the analemma.

Note:

An analemma can be different on different planets. For example, a Martian analemma is different because its tilt and shape of orbit is different to that of Earth's:

Mercury: Because orbital resonance makes the day exactly two years long, the method of plotting the Sun's position at the same time each day would yield only a single point. However, the equation of time can still be calculated for any time of the year, so an analemma can be graphed with this information. The resulting curve is a nearly straight east-west line.
Venus: There are slightly less than two days per year, so it would take several years to accumulate a complete analemma by the usual method. The resulting curve is an ellipse.
Mars: teardrop
Jupiter: ellipse
Saturn: technically a figure 8, but the northern loop is so small that it more closely resembles a teardrop
Uranus: figure 8 (Uranus is tilted past "sideways" to an angle of 98 degrees. Its orbit is about as elliptical as Jupiter's and more elliptical than Earth's. )
Neptune: figure 8

For more, read this.


Is Oumuamua tumbling? given that everyone wonders whether there is a chance that there is something to oumuamua that is not natural, what about tumbling (and/or aiming?)?

is there a natural gravitational or other force that would naturally stabilize non-orbiting asteroids in space?

I would guess that there is a chance that we could detect tumbling.

Assuming that this is a serious question... yes, Oumuamua will be tumbling, just like every other small body.

That it comes from outside the Solar System does not change that it will have three axes of rotation because it is small and thus has not accreted enough matter that it would stabilise into a single primary axis. (This doesn't preclude one of its axes being a primary spin axis.)

And just like every other small body, if we collect enough data in the form of reflected sunlight from its surface, we can construct shape and rotation models and estimate its size and composition.


What causes horizontal and vertical lines coming out of pictures of stars? What causes the horizontal and vertical lines coming out of pictures of stars, instead of them simply appearing as circles? For example, this picture from Wikipedia: .

They are called diffraction spikes, and they're artifacts from a supporting structure inside a reflector-type telescope.


Align multiple fits images without WCS for future aperture photometry analysis Recently I have got a bunch of raw fits images (2048x2048 pixels) of the same sky region without WCS to process in iraf. However, those images are not well-aligned. There would be around 4~5 pixels' deviation for a specific star, thus giving me choppy batch photometry analysis result. Are there any ways to add WCS coordinate using several reference stars or create a user-defined coordinate system so that the same star would have uniform coordinates?

You can use astrometry.net to get the exact WCS of your image.

From the website:

  If you have astronomical imaging of the sky with celestial coordinates you do not know—or do not trust—then Astrometry.net is for you

There is a web application where you can just upload your image and it returns accurate calibrations.


Why can't a quasistar exist now? From my research, I found out that quasistars theoretically existed because of a black hole core whose radiation pressure counteracted gravity within the star. However, a few websites stated that quasistars do not exist now because there are metals contaminating hydrogen and helium.
Could somebody explain to me why metals (or a small displacement of hydrogen and helium) would influence a black hole's radiation pressure, or is it just that there's no where that that much mass (1000+ solar masses) can exist anywhere in one place now?
I looked on Wikipedia (I usually do this and then research on different websites if I find something interesting there) and here1.

Gas clouds with masses much higher than $10^3\,M_\odot$ are plentiful in galaxies; the typical star-forming cloud (the so-called molecular clouds) have masses of $10^3\,M_\odot$ to $10^7\,M_\odot$. When quasistars (hypothetical stars powered not by nuclear fusion, but by accretion onto a central black hole) cannot exist today, it is because all gas in the Universe has become polluted with metals.

Stars form from collapsing gas clouds. In order for a region of a cloud to collapse, it must be sufficiently dense, and sufficiently cool; if it's too dilute, there's not enough gravity, and if it's too hot, the energy of the individual atoms counteracts the collapse, making the atoms escape.

Jeans' mass

This criterion is captured in the Jeans instability equation. The relation can be expressed in several ways; one way is to say that the mass of the cloud — or a small region of it — must exceed the "Jeans mass":
$$
M_\mathrm{cloud} \gtrsim M_J \simeq 3\times10^4 \frac{T^{3/2}}{n^{1/2}}\,M_\odot,
$$
where $T$ (in $K$) and $n$ (in $\mathrm{cm}^{-3}$) are the temperature and the number density of the gas.

From this equation you see that the cooler the gas is, the smaller the threshold. In other words, the smaller stars you can form. If the gas is not able to cool, only the largest clumps will collapse, and hence such stars will be very massive.

Gas cooling

So, how does the gas cool? Hot gas means that the particles have large velocities. If the particles collide, they may excite each other, bringing an electron to a higher state at the expense of slowing down — i.e. cooling. When the electron de-excites, a photon is emitted, which may leave the system. Thus, the kinetic energy of the atoms is converted into electromagnetic energy which escapes.

However, an electron is only excited if the energy of the collision matches closely the energy needed for the excitation. If the collisional energy is too high, or too low, the atoms simply bounce off of each other, maintaining their total energy (although one may transfer some energy to the other).

The effect of metals

If the gas consists only of hydrogen and helium, there are only a few available energies for excitation. Hydrogen happens to be able to cool efficiently around $T\sim10^4\,\mathrm{K}$, while helium cools efficiently around $T\sim10^5\,\mathrm{K}$, but at other temperatures, the gas tends to stay at its given temperature.

However, as soon as there a some metals, the many electrons of these metals, with their many possible transitions, allow for atoms with many possible energies to be excited. Thus, before a gas cloud of $M\sim10^3\,M_\odot$ collapses to form a $10^3\,M_\odot$ star, it will fragment into smaller pieces, forming smaller stars.


Difference between quiet, enhanced and active networks For my research about the sun activity I need to process pictures of the sun. I want the pictures to show 3 stages the quiet, enhanced and active network. I couldn't find any information in the internet about how these 3 networks actually look and how they differ from each other. Can somebody explain this to me? What is showing on which picture?

I'm not sure what you mean by "network" in this context, but on the SOHO website, you can find pictures of the Sun for every day since 2006.

Quiet: there aren't any visible structures at the surface of the Sun

Active: the black spots are sunspots


Which way does the Earth tilt? On which side does the earth tilt to? I have tried figuring it out myself but I can't seem to find the answer.

The equator (Earth's rotation plane) and ecliptic (Earth's solar orbit plane) cross at the vernal equinox ♈, the zero point of both right ascension and ecliptic longitude.
Relative to the ecliptic, the Earth's north pole is tilted toward ecliptic longitude λ=90°, where the Sun appears at the June solstice.
Constellations in that general direction are highest around midnight in December and January: Auriga, Orion, etc.
The Earth's south pole is tilted toward ecliptic longitude λ=270°, where the Sun appears at the December solstice.
Constellations in that part of the sky are highest around midnight in June and July: Sagittarius, Ophiuchus, Hercules, etc.
Even though axial precession changes the celestial poles' orientation relative to the stars in a cycle of about 26,000 years, their ecliptic longitudes with respect to the equinox of date are constant.
The north and south ecliptic poles precess by much smaller amounts and have constant right ascensions α=18h and α=6h respectively, relative to the equinox of date.
Stellarium might help you visualize all this.
In the View window, Markings tab, you can ask it to plot the equator, ecliptic, and celestial and ecliptic poles of date.


How to determine period of pulsar? I've got a file which contains data about photons coming from some pulsar. For each photon I know: a) time, when it was registered, b) a probability that this photon came from pulsar but not from galactic background, c) energy of a photon. My task is to find a period of pulsar.
I suppose I have to use FFT to find a period but there are two issues. First of all, there are lots of photons with low probability (background photons), so I have to somehow select relevant photons. Second, I've got an unequally spaced data, so there is a problem with using FFT.
I am a novice in data processing. Could you please give me links to some lectures or books where this topic is discussed clearly?

Naively, I would say that you're looking for a periodicity in the luminosity.  What about multiplying the probability with the photon energy, and binning those data?  That would allow you to use an FFT.


What is the line of light in galaxy cluster MACS J1206? I noticed this in a thread from world building, what is the continuous line of light in MACS J1206 slightly right of center in this picture? The only thing I could think of would be the side view of a galaxy, but it's much longer than any other galaxies pictured.

I have highlighted the feature I believe you are talking about. It looks like a galaxy that has been repeated (smaller green oval). There are several foreground objects that appear to be causing additional gravitational lensing (red arrows). Due to the shapes of the objects causing the lensing effect, the apparent image can be affected in many different ways, such a image duplication (multiple identical images of a galaxy, etc.), magnification, distortion, etc. The universe is massive and three-dimensional and we still don't understand that much of it yet.


ECEF to ECI conversion How to perform the conversion of (r,v) vector from ECEF to ECI (J2000) coordinate frame?

What I tried

I multiplied the matrix ([cosA -sinA 0] [sinA cosA 0] [0 0 1]) by the state vector, where A is Greenwitch mean sidereal time.

This equation shows that the Z axis doesn't change. However, I have the ephemeris of a real satellite in both coordinate frames, and there Z axis differs!
Where is the issue?

A reference frame that has its Z axis fixed with respect to the Earth's rotation axis is not inertial. The Earth's rotation axis undergoes precession and nutation, and on a shorter term, undergoes polar motion as well.


Brightest star of the night sky I read in many astronomy sites that Sirius is the brightest star of the night sky and that Orion’s Belt points to Sirius. For some days I have been regularly watching Orion (just before dawn) but the star that the Orion's belt points towards is of very average brightness; I don't think it is Sirius. On the other hand I found a very bright star near Orion , albeit not along the line of its belt. On the morning of October 9 (around 5:30 am) this star was adjacent to the crescent moon and shining slightly more brightly ! Even in the early morning glow when all the stars faded away it could still be seen along with the moon. Is this Sirius?

I live in Delhi so I am talking of the Delhi night sky.

The best way to find a particular star/planet/etc. is almost certainly to use a star map of some sort.

For example, doing a Google Image search for "star map sirius" would get you quite a few pictures such as the one in this article about Sirius that give a better picture of how Orion's belt "points" to it.

Alternately, there are a number of computer programs that offer a real time views/maps of the night sky.

Also check out Google Sky.


Can you view the Perseid meteor shower anywhere on Earth? The peak viewing dates of the 2016 Perseid Meteor Shower is said to be between the dates of August 11 through the early morning hours of August 12.

Being a resident of the United States, I keep getting my peak viewing dates and times for my area and for the whole of the United States.  There is never any mention of any other parts of the world having the capability to view this meteor shower.

Is the Perseid Meteor Shower visible to all parts of the world?

You can see perseid meteors from all the Northern Hemisphere, and the Northern part of the Southern Hemisphere. Provided the sky is clear, the moon has set, and Perseus is above the horizon you will see meteors. This shower has a wide peak, so you can see them over several nights.


Where to get Polaris Right ascension value from To calculate for example the polaris hour angle I use this formula to calculate polaris hour angle.

τ = θ - α

τ= hour angle

θ= sideral time

α= polaris RA (~ 2h32m)

But where does the α=2h32m come from? Is there a list? Does this change somehow over time?
Is my understanding correct that this is the time where polaris crosses the upper meridian at Greenwich?

There are several published star catalogs that list right ascensions for stars, along with a lot of other almanac-type data.  Wikipedia's a good resource for common stars (e.g. Polaris).

Right ascensions change over time, as stars move through space.  Star catalogs will have a year associated with them, which is the year that the measurement was taken.


Are harmonic relationships among the rotation periods of the planets well known? Are they factors in other planetary phenomena? Every 39.88 minutes 10 degrees ($\frac{360^\circ}{36}$) of arc passes over the zenith. It can also be stated that the 36th harmonic of the rotation period of the earth (1435.8 min) is equal to 39.88 min. Note the similar relationships with the rapidly rotating planets:

Mars, rotation period of $\frac{1477.2 min}{\text{harmonic of }37} = 39.92$
Jupiter: $\frac{595.8}{15} = 39.72$
Saturn: $\frac{639.6}{16} = 39.98$
Uranus: $\frac{1035}{26} = 39.81$
Neptune: $\frac{966.6}{24} = 40.28$

For the slowly rotating planets, and the Moon, the numbers are larger, but still very close to 39.88:

Mercury: $\frac{3520.2}{88} = 40.00 \text{     } (88 = 22 \cdot 4)$
Venus: $\frac{14582.4}{368} = 39.63 \text{     } (368 = 23 \cdot 16)$
Pluto: $\frac{9223.2}{232} = 39.76 \text{     } (232 = 29 \cdot 8)$
Moon: $\frac{39343.68}{992} = 39.66 \text{     } (992 = 31 \cdot 32)$

The questions are: Are these relationships well known? Are they factors in other planetary phenomena?

I'm very sorry for you, but this is just (mathematical) probability at work and no (astronomical) pattern. I'm on mobile right now so I can't do a full analysis, but note that if you divide Mercury's period (3520.2) by 89 instead of 88, the result is 39.55. Still pretty close to 39.88. In other words, the larger the numbers get, the easier is to find a 'divisor' close to 39.88.


Why is the sunset at different times at the same longitude at equinox? I happen to live in two places that are literally at the same longitude but several thousand kilometers apart from each other (~47°N and ~68°N). It's just a few days until equinox - and I assumed that the sun would set at the very same time in both places. Equinox is supposed to be at September 22, 13:31 UTC. Using suncalc.org I calculated sunset for the northern location at 19:00h while the sun is supposed to set at 18:55 at the southern location.
Can someone please explain why this is?
I can't think of any effects due to refraction of the light as the angle is the same. The only idea I have that it might be due to the definition of sunset. If it is not calculated using the center of the sun, it might be off as sunsets take much longer in polar regions. Still, I'd be surprised if it was 5 minutes.
Also, I do understand that equinox is at 13:31 UTC which does not necessarily align with the sunset (in fact sunset is 3.5h after equinox). But to make things even more strange, even the next day the sun sets later in the north than it does in the south - I'd expect it the other way round. It is actually another day later, September 24, that the sun will set at exactly the same time in both places - (more than) two days after equinox.
PS: Not an astronomer myself, I'm just curious - please bear with me...

The only idea I have that it might be due to the definition of sunset.

The instance of sunrise occurs when the first bit of the Sun becomes visible at the horizon, while the instance of sunset occurs when the last bit of the Sun disappears below the horizon. This means the actual position of the center of the Sun is about 50 arc minutes below the horizon at sunrise / sunset.
At sunrise and sunset on the days of the equinoxes, the angle between the Sun's path and the horizon is roughly equal to 90° minus the absolute value of an observer's latitude. The Sun's path is much more oblique at your more northern location.
It takes the Sun about 200 seconds (3.333 minutes) to traverse those 50 arc minutes at the equator on the equinoxes. At some latitude $\lambda$, it takes about $200 \sec(\lambda)$ seconds for the Sun to go from 50 arc minutes below the horizon to the horizon on the days of the equinoxes. That's 4.89 minutes at 47° latitude and 8.90 minutes at 68° latitude, a difference of four minutes.
The remaining minute could easily be due to rounding. Web sites typically publish sunrise and sunset times rounded to the minute because the difference between actual and predicted times varies by over a minute due to due to variations in the atmospheric state.


How to configure Celestia to work with a spherical mirror projection on a dome I find hints around the web, e.g. http://www.cosmodome.net/mirrordome.php that Celestia can be projected on a dome using a spherical mirror projection configuration.

The problem is that I can't find any documentation on how to configure Celestia to warp the generated image, so it can be used in such a projection setup.

For Stellarium\Nightshade I found some references, e.g. from spherical projection guru, Paul Bourke, and I expected to find similar resources for Celestia.

Can Celestia be configured for a spherical mirror projection system?
Can you point me to documentation on how to set up Celestia for a spherical mirror projection system?
Is there a way to warp Celestia with 3rd party tools and then project the images?

According to Paul Bourke's page on screen capture

  While only the commercial version of Celestia supports fisheye, it also does not support warping of fisheye (although some versions have been created that do). The solution outlined here has been tested successfully with a fisheye supported version of Celestia.

It seems that Celestia in its free version is not providing any support for dome projection and the commercial version requires tweaking before you can use its support for fisheye projection and convert it to Spherical Mirror Projection.


Do stars vary their own brightness? All I could find on the internet was about how stars vary in brightness depending on their distance to Earth, temperature, type of star...

But my question is, can a star can change its brightness.

They can, and some do. These stars are called variable stars, because their luminosities as observed from Earth vary over time, often (though not always) in a regular period. Here are some broad categories:

Pulsating stars, where fluctuations lead to increases and decreases in size or temperature, which in turn produce changes in the star's brightness. Example: The oft-cited Cepheid variables.
Eclipsing binary stars involve two stars orbiting each other. When one passes between the other and Earth, the combined luminosity of the system appears to decrease, even though the stars' intrinsic luminosities are probably constant. Example: Algol.
So-called eruptive variables (often lumped together with cataclysmic variables, which are different) may have irregular outbursts caused by flares or other phenomena. Example: Luminous blue variables (LBVs).

The change in brightness and the length of the variations depend on the type of variable star in question. The luminosity can vary by anywhere from a fraction of a magnitude to many magnitudes. Likewise, the period (if there is one) can range from hours to years.


If the hammer and feather move at the same speed why do comet and the tail particles move at different speeds? From this exhibition by an astronaut on the moon dropping a hammer and a feather and showing they move at the same speed, why do the particles of the comets tail move away from the comet, slow down and fall away as it disintegrates? Shouldn't they move at the same speed of the comet?
For example if the feather was attached to the hammer and it was falling and broke off wouldn't it fall at the same speed, not start to trail behind and eventually fall away?

The tail of a comet is not actually "slowing down and falling away" from the comet, like you might expect to see when smoke streams out from behind a moving object on earth. The tail of a comet is actually being pushed away from the sun by the solar winds and radiation. That's why the tail of a comment always points away from the sun, and doesn't stream out behind it.

If there were no other forces at work, the coma would actually travel at about the same speed at the comet. As the material surrounding the core of the comet vaporized, it would form a coma ("atmosphere") made up of sublimated gas mixed with dust, moving along  at the same speed (more or less(1)) as the comet… like you observed in the video.

(1) In reality, without the solar winds, the coma of the comet would continue to move outward, as its expulsion velocity would easily overcome the comet's weak gravitational pull.


is there a way to determine a distant galaxy’s speed relative to the Hubble Flow by measuring time dilation effects between there and this galaxy? As I understand it there is a preferred frame of reference based on the velocity for the CMB and our galaxy is moving in relation to it (at about 600 km/s). I think this is how the Hubble Flow is defined. Given that there are no relativistic effects, and I really mean time dilation, between distant galaxies that are moving apart due only to Cosmological expansion, is there a way to determine a distant galaxy’s speed relative to the Hubble Flow by measuring time dilation effects between there and here? If so using that and observed red-shifting, one could establish a more accurate measurement of the Hubble Constant.

No. Time dilation is expected and observed as a result of universal expansion (e.g. in the light curves of type Ia supernovae and the duration of gamma ray bursts, Blondin et al. 2008; Zhang et al. 2013). There is no way of distinguishing between a Doppler shift and cosmological redshift for an individual source.


Which observatories help NASA find 5 new asteroids per day? I just read the following in the NPR website news article NASA's New 'Intruder Alert' System Spots An Incoming Asteroid.

  NASA pays for several telescopes around the planet to scan the skies on a nightly basis, looking for these objects. "The NASA surveys are finding something like at least five asteroids every night," says astronomer Paul Chodas of JPL.

Which telescopes are the "several telescopes around the planet to scan the skies on a nightly basis" that NASA pays?

Also, is it really roughly five new actual asteroids identified per day (almost 2,000 per year), or is that 5 objects detected, and only some small fraction turn out to actually be unique, new objects?

5 per day sounds like a lot.

Given that the article seems to be referring to NASA's Near Earth Object Program, it appears that there are five subprograms scanning the skies:

Catalina Sky Survey (CSS) (four telescopes: 0.50 m, 0.68 m, 1.00 m, 1.50 m)
Pan-STARRS (one telescope: 1.80 m)
LINEAR (two telescopes: 1.0 m (retired), 3.5 m)
Spacewatch  (two telescopes: 0.9 m, 1.8 m)
NEOWISE (one telescope - WISE: 0.4 m)

According to JPL, CSS and Pan-STARRS are responsible for 90% of new Near Earth Object (NEO) discoveries. LINEAR is responsible for detecting objects one kilometer in diameter or greater, while NEOWISE provides additional measurements from previous WISE targets, so it isn't making new detections, merely follow-up observations.

NASA's statistics show that over 15,000 Near-Earth Asteroids (NEAs) have been discovered so far, from all programs scanning the sky. In each of the past three years, about 1,500 NEAs have been discovered, most by the five programs listed above. That's roughly consistent with the five-per-day claim. However, it's not yet near 2,000 per year.


Calculating Radial Velocity from Hipparcos Using data from Hipparcos is it possible to calculate Radial Velocity.  If we take 19 Leonis as example, RV is -3.70.  Is it calculated from Proper Motion?  Can I ask that any formulas be done in long hand please, i.e. how you would enter it into a standard spreadsheet rather than with symbols etc.

http://simbad.u-strasbg.fr/simbad/sim-basic?Ident=19+Leonis&submit=SIMBAD+search

Additionally

Took the advice and found CRVAD2.dat ( http://cdsarc.u-strasbg.fr/viz-bin/Cat?III/254 ).  The following line comes from CRVAD2.dat file and when I looked up HipId 40829 which is below, the RV value is -3.70 which matches the link above.

9.790554129  11.56817726   0.8   0.1   11.06   1.00   -48.63    -3.39   1.24   0.69  6.709  6.436 0.005 0.005 0.007   D   AY D A3n          A 7 VN                48029  84722  948942 4 ABA    -3.70  -9.99 B   9 1

If I look up Sirius (32349), it says the RV is -7.60 but when you look up Sirius at Simbad, it says -5.50 ( http://simbad.u-strasbg.fr/simbad/sim-basic?Ident=Sirius )

6.752569403 -16.71314306   1.2   0.1  379.20   1.58  -552.63 -1206.40   0.50   0.50 -1.430 -1.121 0.011 0.002 0.039   U     OD A1V+A5wd     A1V+A5wd              32349  48915 1447791 4 AB     -7.60  -9.99 A   0 1

Whats the reason for the discrepancy, CRVAD2.dat is an old version like there's two Hip data files (97 and 07)?

You cannot calculate the radial velocity from a proper motion. The proper motion is the motion you see the star move on the sky, so perpendicular to your line of sight towards the star. The radial velocity is the velocity of the star along this line of sight. Perhaps this picture will help:

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

Radial velocities are actually calculated by measuring the Doppler shift of the spectra of stars, but this is not very easy, and you need the raw spectrum data. You are probably best off just taking the values of the radial velocity from that site for example.


Pair together light curve and radial velocity data of specified star I would like to create a web application that works with light curves and radial velocities of stars with exoplanets. I found a NASA bulk data download with both light curves and radial velocities (it's exactly what I'm looking for). However, I don't know how to pair together the light curve and radial velocity of a single star (without downloading the whole dataset).

For example, using the script below I can download the light curve of KIC 10666592, where there is 1 exoplanet:

wget -O 'kplr010666592-2009131110544_slc.fits' 'http://exoplanetarchive.ipac.caltech.edu:80/data/ETSS//Kepler/005/755/19/kplr010666592-2009131110544_slc.fits' -a search_33681064.log

Now, how can I download radial velocity data for this specific star? Scripts for radial velocity look like (there is no KIC in URL):

wget -O UID_0000522_RVC_001.tbl http://exoplanetarchive.ipac.caltech.edu:80/data/ExoData/0000/0000522/data/UID_0000522_RVC_001.tbl -a RADIAL.log

Is there any way to connect KIC (Kepler input catalog) and radial velocity UID?

An excellent page to get most or all names for a star is Simbad which also happens to know the Kepler IDs (KID).

Ont the other hand, every of the RV data files contains info on the star's identifier it belongs to in the header, e.g:

  \STAR_ID='HD 4628'

So using Simbad you should be able to do a cross-matching. SimBad allows scripted query and includes all alias names for stars; thus query for one known star name and look for identifiers found in the list of RV-observed stars or vice versa. Simbad allows to be queried via scripts, so you can make a short programme to do this task for you.

However I'm not 100% convinced that you necessarily find many matches between those two data sets; it's easier to do photometry than spectroscopy.


Can we have a second moon? This question sounds like silly and rather fantasy filled. But can our earth's gravity capture a sizable asteroid and make it rotate around our earth like our moon which become visible from ground. Is there any theoretical possibility or practical probability?

At this stage of our solar system it is impossible for something with size of moon to fall in inner orbit. But smaller things like asteroids do fall in, and become captured by earths gravity. but they aren't big enough to see with naked eye. For example 2006 RH120 is a near earth asteroid which orbited Earth from September 2006 to June 2007.


redshift in astrophysical plots Why do astrophysicists often use redshift in plots? Why it is important to know how an astrophysical object evolves in function of redshift? I was seeing an "cosmological time per redshift galaxies formation plot" and thinking why redshift and not, who knows, density?

Redshifts are plotted because usually that is all that can be measured.
Other quantities could be plotted, but since these are inferred from the redshift itself (e.g. density would just mean plotting $(1+z)^3$), or would need to assume particular cosmological models or parameters (e.g. if you plotted look-back time), then it makes sense to stick with plotting the observed data.


Why can't neutron stars ignite and explode? Beyond the Chandrasekhar limit, white dwarfs become extremely hot. As a result, previously unfusable carbon can become fusable, causing nuclear reactions. This leads to a thermal runaway and ultimately a type Ia supernova.
However, this doesn't seem to happen in neutron stars. They can keep gaining mass until they become black holes. After a certain point, shouldn't the immense temperatures cause a thermal runaway like in white dwarfs, leading to their explosion?

In a white dwarf, the dense matter is not in its lowest energy configuration. Energy can still be extracted from the white dwarf material by fusion, provided it can be ignited.

What exothermic nuclear reactions would there be that could take place in a neutron star? The bulk of the material is in the form of neutrons with a small number of protons and electrons. At these densities, that is the most stable equilibrium composition.

If a neutron star gains mass in a gradual way, then the most likely course of events will be that its radius will decrease (that is what happens in objects supported by degenerate matter) until it reaches a General Relativistic instability where its collapse to a black hole is inevitable (when $R$ is somewhere between 1.25 and 2 times the Schwarzschild radius). It is possible that neutrons may transform before that into additional hadronic degrees of freedom or into quark matter, but these are endothermic processes that suck kinetic energy out of the neutron gas and only hasten the collapse.


Is a planetary system star's referred to as their sun? It's my understanding that the Sun (uppercase) is used to refer to our sun/star, because it is The Sun. Latin name is Sol, hence the Solar System. Even the tag (lowercase sun) on this post has the caption "Questions regarding the closest star to Earth, at the centre of the Solar System."

Other planetary systems have their own stars, which I understand are called a sun, in the context of that planetary system.

I've tried my best to find a credible source that exoplanets indeed have their own suns, and that the term "sun" doesn't just refer to our Sun. I have only found this one website that backs this up: http://curious.astro.cornell.edu/physics/78-the-universe/stars-and-star-clusters/general-questions/346-what-is-the-difference-between-a-star-and-a-sun-beginner

It seems to be common knowledge in the scientific community that this is correct, but I can't find anything else online to back this up. Help?

The technical name (in English) for the Sun is not Sol, which is just Latin for sun. The technical name for the Sun is the Sun. Another body in the sky has a similarly boring name, the Moon. There's one more boringly named object: in the Solar System: The Earth. Note the use of "the" (a definite article) and the use of capitalization to indicate a specific object.
The problem with Sun, Moon, and Earth is that we have been using these names (or their predecessors) for thousands of years. For example, sun, sol, ἥλιος (helios), and a bunch of other names for that very bright object in the sky whose presence distinguishes day from night all derive from the same proto-Indo-European word sóh₂wl̥.
As an end note, from http://curious.astro.cornell.edu/about-us/159-our-solar-system/the-sun/the-solar-system/4-what-are-the-names-of-the-earth-moon-sun-and-solar-system-beginner,

You may read or hear people using Luna for the Moon, or Terra or Gaia for the Earth, or Sol for the Sun, but in English-speaking countries, these are poetic terms, often seen in science fiction stories, but not used by astronomers in scientific writing. In some countries where Romance languages are spoken, these terms are the official names.
It's also interesting to note that most astronomers do not call our galaxy the Milky Way in technical writing--they call it the Galaxy.