Orbital decay via gravitational emission in a hypothetical, two-body universe Assume we have a universe with nothing in it except for two objects. One of them is a stationary object that is perfectly spherical with the mass of the Moon (without any atmosphere, of course; also, let's call this Body A). The other body is a small satellite similar to the Hubble Space Telescope, orbiting at 100 miles (161 km) above the surface of Body A.
How long would it take for the satellite to crash into Body A via gravitational emission?

Using this equation from Wikipedia for the approximate merger time:
$$\frac{5}{256}\frac{c^5}{G^3}\frac{r^4}{m_1m_2(m_1+m_2)},$$
and Wikipedia's values for the Moon's equatorial radius and mass, and the launch mass of the HST, we can use this query in Google:
((5/256) * (c^5/G^3) * ((100 miles) + (1738.1 km))^4 / ((11110 kg) * (7.347631E22 kg)^2))
to perform the calculation.
The result is $\approx 3.4497 \times 10^{46}$ seconds, or around $1.093 \times 10^{39}$ years, which is about $7.93 \times 10^{28}$ times the current age of the universe.
That equation is for a circular orbit, and it's only a crude estimate based on orbital energy lost via gravitational radiation. It mostly ignores the change in the orbital radius, since that's fairly small until the final stages of the merger.
There's some information about the effect of orbit eccentricity  on changes to orbital angular momentum and energy due to gravitational radiation at Two-body problem in general relativity.


Why isn't the star that created the black hole a black hole? If the mass of a black hole is creating so much gravity that light cannot escape, why isn't the mass of the star that created the black hole (before it went supernova) trapping light as well?

By all accounts, that pre-supernova star should have boatloads more mass than the black hole post supernova, right?  Doesn't the star lose most of it's mass when going supernova?

You are correct in saying that a star loses a lot of its mass in a supernova. However, there is a reason why the star still becomes a black hole. Actually, I suppose the question here is "Why doesn't a star become a black hole before it even undergoes a supernova?"

There is a reason for a supernova (I'll assume you're talking about type II supernovae, which result from incredibly massive stars). Stars undergo nuclear fusion, and this leads to "thermal pressure", which counteracts the force of gravity. Without this pressure, gravity would indeed make a sufficiently large star collapse upon itself. Gravitational collapse occurs when there is not enough pressure to counteract gravity; the result is a spectacular supernova. So stars only become black holes (or other compact objects, such as neutron stars) when they cannot produce enough energy to counteract the force of gravity due to their own mass.

As for the first part of your question (sorry for answering in reverse), light in the area of a black hole cannot escape if it is inside its event horizon or on a trajectory towards it. The radius of the event horizon for a non-rotating black hole is its Schwarzschild radius, which is proportional to the mass of the black hole. The reason this is non-applicable in stars is because the Schwarzschild radius in stars is deep inside its interior, and there is not a strong enough gravitational field to produce an event horizon to trap any light near it.

Thermal pressure reference: https://en.wikipedia.org/wiki/Gravitational_collapse

I hope this helps.


Is "Main Sequence" a temporal sequence? Stars plotted by luminosity and surface temperature fit into patterns in a Hertzsprung–Russell diagram.  A rough diagonal subset of this plot is called the main sequence.  Is this in any sense a temporal sequence?  There's a clue in the stellar physics section of the wikipedia article that the answer is no, but that it once was thought so:

  Contemplation of the diagram led astronomers to speculate that it might demonstrate stellar evolution, the main suggestion being that stars collapsed from red giants to dwarf stars, then moving down along the line of the main sequence in the course of their lifetimes.

So does the word "sequence" in this case now mean just a particular ordering, and not a progression over time that any one star makes?  Is the main sequence just some kind of plateau in stellar evolution where stars spend a significant duration?

No, the Main Sequence is more like a starting line. Most stars spend a long time on one point of it (10 billion years for the sun) while they fuse hydrogen into helium. Then they wander off it.

In this diagram the black line is the Main Sequence. The colored lines show temporal sequences. This progress through time of an individual star is called its evolutionary track.

The numbers along the black line are solar masses (1=the sun). This diagram evolved through a sequence of images by Wikimedia users Rursus, G.A.S, and Jesusmaiz.


Why is the shadow of the earth not visible during sunsets? I would imagine that during a sunset, for a brief moment of time, there would be a shadow of the earth seen in the clouds but I have never seen this. Why does it not happen?

To have a sharp shadow, you need a point source of light.  But the Sun is extended enough that it does not set all at once, so there is not a sharp shadow.  Also, there is refraction of the redder light, so instead of seeing a shadow, you'd see reddish clouds-- which I wager you see quite often (though much of that effect is from the way the atmosphere scatters away the blue light-- those effects are hard to disentangle but you know refraction matters because you see how the Sun's shape is distorted upon setting).


Has NASA confirmed earth now has a second moon orbiting the planet? Several sites claim NASA has confirmed a second moon orbiting earth. This is the Asteroid 2016 HO3. One source states The newfound quasi-satellite/moon-2016 HO3, is likely larger than 120 feet (40 m) and smaller than 300 feet (100 m).

http://www.physics-astronomy.com/2016/06/nasa-just-confirmed-that-earth-has-new.html#.V3tdaNIrLIU

Question
Is this claim true circling the internet and can anyone post a link from NASAs offical web site to conclude this is true?

It's not true; this object is not a moon of Earth's.

Here's a NASA/JPL announcement of it. 2016 HO3 is the "name" being used for it at the moment.

They're calling it a "quasi-moon". It's in an orbit that is in resonance with Earth's but is not permanently bound. (It's not the first object to be found with such properties too.)


USNO moon images look quite different? I visited:

http://aa.usno.navy.mil/imagery/moon

http://www.usno.navy.mil/USNO/time/moon-phase-images

within seconds of each other, and saw two very different "current moon images":

Is one of these images wrong? The larger image seems to display a much
bigger crescent than the smaller image.

I've noticed this earlier as well.

This is an unofficial explanation, but I doubt you'll get an answer from official sources here, unless you ask them directly (e.g. their Twitter account is @NavyOceans), so I'll give it a shot. The small thumbs appear to be merely approximation, representation icons that aren't resized from the real-time image, but merely prepared in advance images, displayed depending on the current Moon's phase, calculated by the server.
If you try their Moon phase image generator (at the bottom of this page), you'll see why they did it like that. The prepared thumbs are numbered from m180.gif for Full Moon to m360.gif for New Moon, with a 2° step, and apparently the server calculates based on input parameters the Moon phase and displays prepared thumbnail with a minimum approximation precision of 2° (roughly 8 hour step). The same page also offers an explanation for how this thumbnail generation works:

These lunar phase images were created by R. Schmidt from ray-traced
images of the Moon. A Clementine spacecraft mosaic of the lunar
surface was mapped onto a sphere, and scenes were rendered as a
virtual Sun "orbited" the Moon. The depiction of lunar surface
features suffers geometric distortion but the terminator is correct
with respect to the spherical Moon.

The lunar phase calculator offers date selection that goes forward in time or before the time any images of actual observations were made and stored, so it's unreasonable to expect actual images displayed for, say, some date in 22nd or 19th century. Some examples off U.S. Naval Observatory's server:

(source: navy.mil)

(source: navy.mil)

(source: navy.mil)

(source: navy.mil)

(source: navy.mil)
            Full Moon                        Waning Gibbons                     Last Quarter                    Waning Crescent                     New Moon
The larger image of how the Moon looks like now seems to be generated by a lot more precise renderer (image is refreshed every minute by a JavaScript call) that takes into account not merely lunar phase, but also its libration in respect to the surface on the Earth. It would look more like any frame off this animation:

                         Fig. 2: Lunar librations in latitude and longitude over a period of one month (Source: Wikipedia)
And inclination is likely calculated for geolocation of the Astronomical Applications Department of the U.S. Naval Observatory in Washington, DC. The page however doesn't explain if this assumption of mine is correct, but it would certainly appear so.


(Thought experiment) if we put a huge mirror into space, could we see back in time to the big bang? This is an addition to this question which is closed:

By putting a mirror in space, would we be able to see into the past?

Imagine there is a supernova a hundred years ago. Lots of people see it but nobody has a decent telescope. A hundred years later we point a powerful telescope in the opposite direction, and find a huge alien mirror a hundred light years away, that allows us to watch the supernova.

My question is, what is the limit of such a technique?

Could we see right back to the big bang if the mirror was large enough and far enough away?

Please note this is not a question about whether the telescope or mirror are realistically possible.
Assume we are using some superior alien technology.

Imagine there is a supernova a hundred years ago. Lots of people see it but nobody has a decent telescope. A hundred years later we point a powerful telescope in the opposite direction, and find a huge alien mirror a hundred light years away, that allows us to watch the supernova.

Light travels at a fixed speed so a hundred years later when we build our telescope we would see nothing in the opposite direction, because the light reflected from your alien mirror would only have reached that mirror, and has another hundred years to reach our telescope.

  My question is, what is the limit of such a technique ?

It's a thought experiment, not a technique.

  Could we see right back to the big bang if the mirror was large enough and far enough away?

Light takes time to reach us, so when we look at very distant objects they're already telling us about the distant past.  This is how we have built up an understanding of what happened in the distant past.

You might want to explore the subject of the Cosmic Microwave Background here.

In principle it would be easier to detect any radiation from the big bang directly (without a mirror) in a similar way.  But the universe is not a quiet place and radiation from the big bang has literally been through everything and won't be as easily detected (if it's possible at all).  We don't even have a proper theory to know what to expect to be able to detect, if anything.

All the mirror does is complicate things.


How would one calculate the synodic period of the Earth and an elliptical orbit? For example, when would someone with a telescope be able to see Starman and his Roadster (when will the Tesla roadster's elliptical orbit cross ours again, and how would that be calculated?)

For two circular orbits, I understand this is given by $$\frac{1}{S} = \frac{1}{T_2} - \frac{1}{T_1} $$ .

tl;dr:

The synodic period would be the same. For this object it will be about:

$$\left( \frac{1}{365.25} - \frac{1}{557} \right)^{-1} \approx 1061 \ days.$$

You can confirm that by reading this new ArXiv preprint about Roadster: The random walk of cars and their collision probabilities with planets. In the 2nd paragraph of Section 3 it says:

  The bodies reach the same orbital longitude on their synodic timescale of ∼ 2.8 yrs.

There's a slight difference, since the Roadster's orbit continues to be refined, and the paper is about evolution over long timescales.

However, I think what you really want to know are the times of closest approach when (at least) one orbit is significantly eccentric, and those  will not be periodic or come at regular intervals.

Well the Roadster is quite far away now and moving farther rapidly. See this answer for some details and some nice images from medium-sized tracking telescopes. It's already dimmer than +22 magnitude as seen from Earth, and as far as I know there won't be any really close approaches to Earth for a long time.

But to answer your question, the period is about 557 days. You can see this for example if you go to JPL's Horizons and look at the osculating elements. If you look a few weeks in the future when the Roadster is sufficiently far away from Earth not to be perturbed, the period around the Sun flattens out to 557 days.

above x2: Top: Distance data from JPL's Horizons plotted using Python from here. Bottom: Estimated apparent magnitude using distances from Horizons plus math shown in the text of this answer. Your milage may vary, but probably within +/-2 magnitudes of "official" predictions when they come out.

A really cool GIF can be seen in the Space.com article Observatory Spots Elon Musk's Tesla Roadster Zooming Through Space (Video). It is over 7 MB so I can't add it here. However, here are ten frames from somewhere in the middle. Still, you should go see the whole thing.

GIF:

DEMIOS image below, from here as also tweeted by Jonathan McDowell. The little dot near the center moving to the right and up is Roadster in reflected sunlight, probably mostly from the white FH 2nd stage still attached.

GIF:


Understanding The Turnover Point of Angular Diameter Distance I am trying to get a better understanding of cosmological distances, in particular the angular diameter distance which I have also seen referred to as angular size distance.

What I am looking for is a more detailed explanation of why the angular diameter distance "turns over" in the following graph.

Wikipedia states in their article on angular diameter distance that objects beyond a certain redshift have a smaller angular diameter and as such appear larger on the sky.

I have searched around and have been unable to find a complete explanation of this behaviour and have only found brief answers which state the expansion of the universe or the size of the universe.

I would like if possible for a more detailed explanation as well as the physical significance if any at all of what it means for an object to be at this redshift when they emitted light we are now observing.

Thank you for any feedback or and discussion or suggestions where to look. I would also like to include as it may be of some help that I am currently in my final year of undergraduate studies in physics so please do not hold back any details or technicalities!

On the one hand an object spans a smaller angle the farther away it is, as expected. On the other hand, due to the expansion of the Universe and the finite speed of light, very distant objects were closer to us when they emitted the light we see today. At that time they spanned a larger angle.

The following paragraph is hard to understand without a drawing , so have a look at the drawing in this answer, which may help. The light from the edges of a galaxy that happened to travel toward where we were at the time it was emitted did not reach us, because by the time it arrived to that place, we were no longer there. But the light that happened to be emitted toward some point behind us reached that point at the same time as we did, so that's what we see.

The exact turnover, the threshold between "looking smaller because far away" and "looking larger because closer in the past" depends on the expansion rate history of the Universe, as well as on the way that light propagates in the Universe, which in turn depends on the densities od the various constituents of the Universe. The interrelationship of these quantities is given by the Friedmann equation.


What are the key coordinate systems used in Astronomy, and where are they used? Wikipedia describes the five following celestial coordinate systems each with rectangular/spherical variants:

horizontal
equatorial
ecliptic
galactic
supergalactic

What are the practical preferred uses of each, in relation to the others. If you were navigating within the solar system, which system would you use, or has been used by existing craft?

For example wikipedia describes the ecliptic system as "still useful for computing the apparent motions of the [celestial bodies]", which seems pretty bloody obsolete. Even though it seems to me as the most intuitively useful system.

Navigating the solar system, you'd probably track your probe using Earth-based radio telescopes in the equatorial ICRF, though you might describe the overall trajectory in terms of the ecliptic.

Navigating at sea, you might measure the Sun's position in horizontal coordinates at a time in GMT to determine your own geographic coordinates.

Practical uses in backyard astronomy:

horizontal: telling someone where to look relative to an object on the horizon; pointing a telescope on an altitude/azimuth mount
equatorial: finding or plotting an object in a star atlas; choosing atlas pages to use on a given night; pointing a telescope on an equatorial mount
ecliptic: importing comet or asteroid orbital elements into planetarium software; planets are easiest to observe when their longitude differs most from the Sun's
galactic: low latitudes have more Milky Way star clusters and nebulae; high latitudes have more external galaxies
supergalactic: pondering the distribution of galaxy clusters?


Would the Event Horizon Telescope be able to produce a superior image of Betelgeuse? Current images of Betelgeuse are already amazing, but I was wondering if the EHT could be able to make a significantly better image, given that Betelgeuse is pretty big and pretty bright?

Presumably, yes, EHT observations could improve on existing radio observations of Betelgeuse (e.g. recent ALMA images and comparatively ancient VLA images). Any observations would likely be targeted at known photospheric mm or super-mm emission from the star, mapping radius and temperature changes. The ALMA data showed the existence of a spot $\sim1000$ K hotter than neighboring regions, which would ideally teach us something about convection inside the star.

With a (current) resolution of a couple tens of microarcseconds at $\lambda\sim1.3$ mm, the EHT would indeed improve on existing interferometer results by orders of magnitude (which I believe, in the case of ALMA, are the highest-resolution images of Betelgeuse, period, at any wavelength).


Overhead (galactic North) view of the stars in and around the Vela Constellation with distance I am working on a project and would like to have a star map--seen from "above" (Galactic North?) that calls out the stars in the Vela constellation and shows their distance from sol / earth.

I realize that any 2D map will not show the Z-axis, but I'm okay with that (especially if there's a notation).

I want to be able to show which stars in the Vela constellation are closer to earth vs. further away.

Is there anything like this? I've looked and can't find it.

note: I'm about to practice Astronomy without a license, if I've said something wrong please feel free to comment or just edit and correct it, thanks!

Wikipedia's Vela (constellation) lists its brightest stars and links to each star's Wikipedia page. Here are the first five of them:

https://en.wikipedia.org/wiki/Gamma_Velorum
https://en.wikipedia.org/wiki/Delta_Velorum
https://en.wikipedia.org/wiki/Kappa_Velorum
https://en.wikipedia.org/wiki/Lambda_Velorum
https://en.wikipedia.org/wiki/AH_Velorum

On each page at the bottom of the data box on the right side there is a link to the star's page in Symbad. For Gamma Velorum (HD 68243) for example, the page is http://simbad.u-strasbg.fr/simbad/sim-id?Ident=HD+68243

On that page you can see the galactic coordinates (GAL coor) are about 262.806, -07.697 and since you only want a 2D map use the azimuth (262.806) to draw a line from our position in the solar system, which is the origin of galactic coordinates (for more on that see answers to How was the galactic plane established? and also can you translate a GPS coordinate to a Galactic coordinate.

Now how far away is the star from us on that line? Use the parallax value for that. For the Gamma Velorum example it is about 3.556 milliarcseconds. the distance to an object (measured in parsecs) is the reciprocal of the parallax (measured in arcseconds). So the distance here is 1/0.003556 or 281.2 parsecs which is about 917.2 light years.

For your 2D projection on to a flat map, you'll need to foreshorten that distance, so multiply by cos(altitude): $\cos(-07.697)=0.9910$ so you put your dot along the azimuth line at 917.2 times 0.9910 = 908.2 light years.

You'll have to decide what to do if one of the stars is far away from the plane of the galaxy, because the projected distance won't accurately reflect the actual distance.


Do white dwarfs lose mass as they fade to black dwarfs? Is there a correlation between temperature, mass, and radius? I'm having a difficult time understanding certain behaviors of white dwarfs.

I understand how mass is lost in the red giant to white dwarf transition process. I understand that white dwarfs can accumulate mass from a partner binary or some other source and I'm aware of the 1.4 M☉ Chandrasekhar limit. I get that they're not losing all their mass; in general they're just cooling and in the end there will be a lone black chunk of crystallized material floating forever in the void. I know they are really insanely hot when formed and very cold after an unfathomable amount of time. But is it that same, initial mass, or does it somehow diminish?

But is that initial mass the mass the remnant has after its cooled to a black dwarf? I guess what I'm asking is that I'm trying to figure out if there is a correlation between temperature, mass and radius. Some sort of equation?

The mass of a white dwarf continues to be that which it was "born" with. It will not change significantly unless it accreted material from a companion.

The radius of a white dwarf is, to first order, given by the "mass-radius relationship" and this relationship does not involve the temperature.

The mass-radius relationship is appropriate for a "cold" star. "Cold" in this context mean that the pressure that supports the white dwarf is only dependent on density, which is the case for the degenerate electrons in the interior, which have kinetic energies much greater than their thermal energy.

However, degenerate electrons also have excellent thermal conductivity, so white dwarf interiors are isothermal. Yet, they have a density gradient - denser in the middle and less dense answer move outwards. At some point close to the surface the electrons are hot enough to stop being degenerate and the gas pressure becomes temperature sensitive.

What this means is that the radius of a hot white dwarf is bigger than that of a cold white dwarf of the same mass. The effect depends on both the mass of the white dwarf (bigger for lower mass white dwarfs with lower surface gravities) and its age (since white dwarfs cool as they get older).

This effect is not negligible and needs to be properly modelled in order to understand the luminosities of white dwarfs.

The plot below is from Parsons et al. (2017) and shows (as points) measurements of masses and radii of white dwarfs in eclipsing binary systems. The lines are model curves for white dwarfs with surface temperatures ranging from zero (dashed line) to 60,000 K (appropriate for a very young white dwarf) in steps of 10,000 K. Clearly, there is not a unique mass radius relationship and the radii do apparently depend on temperature (and core composition), as the model curves suggest.


Has Hawking Radiation Ever Been Observed? I know Hawking Radiation has firm theoretical footing, but has a signal ever actually been observed? What observational research is being done to attempt to see this effect? Is it simply too tall an order (i.e. - too far away, too weak of a signal, etc..)?

According to this MIT Technology Review "First Observation of Hawking Radiation" and an associated article "Hawking radiation from ultrashort laser pulse filaments" where the authors have simulated Hawking radiation in the laboratory, using ultrashort laser pulse filaments.

So, in part answer to your question, yes Hawking Radiation is believed to have been observed, but in a laboratory environment.

According to "Observing Hawking radiation in Bose-Einstein condensates
via correlation measurements" (Fabbri, 2012), direct observation of the Hawking radiation is (in their word): hopeless.

However, once again simulations are possible:

  Hawking effect, which depends only on kinematical properties of wave propagation
  in the presence of horizons, is present also in nongravitational contexts,
  for instance in stationary fluids undergoing supersonic flow.

From these articles, it would seem that direct observations of Hawking radiation is not feasible currently; however, we are able to produce analogues of the phenomena in the laboratory.


Question about radiation pressure A spaceship with mass $m = 25\:kg$ is moving toward a red giant with $M = M_{sun}$ and $L = 1000\:L_{sun}$ (L: Luminosity). It is moving on a line (so this is not an orbital question). When it gets too close to the red giant, it's sensor open its solar sails (they get and then reflect the light completely), hence the radiation pressure help it to escape from the red giant (on the same line). Find the area of solar sail. (It is from National Olympiad of an Asian Country)

My problem is with the radiation pressure formula. for solving this problem, somewhere said the radiation pressure is: ($c$: light speed, $\sigma$: Stefan-Boltzman Constant)

$$P= \frac{4\sigma T^4}{3c} = \frac{4}{3}\frac{L}{4\pi D^2}$$
Where $T$ is the equivalent Temprature in distance $D$.

Somewhere else said:

$$P= \frac{4\sigma T^4}{c} = 4 \frac{L}{4\pi D^2}$$

In another book it is:

$$P=2 \frac{4\sigma T^4}{c} =  2\times 4 \frac{L}{4\pi D^2}$$

2 is because it gets the light and then reflecting it.

At the final:

$$P\times A = \frac{GmM}{D^2}$$

I want to know what is the right solution for calculating radiation pressure?

I must say that the below picture is also with the question but in the books, it was used for another part of the question which is not my problem and I didn't mentioned it above.

I might venture into some math which is slightly above what you need, but I wanted to show you the reason behind the correct equation to use. To start with, the pressure due to radiation, in the most general sense, is given by

$$P_{\mathrm{absorption}} = \frac{\langle S\rangle}{c} \cos(\theta)$$
$$P_{\mathrm{reflection}} = 2\frac{\langle S\rangle}{c} \cos^2(\theta)$$

The top equation represents the case where some light hits your object and the light gets absorbed, and the bottom equation represents the case where light hits and bounces off your object (really it's re-emitted). Your case is the bottom equation so I'll work with that from now on and drop the reflection subscript. Without needing to explain exactly what it is, just accept that $\langle S\rangle$ is the average energy of the photon being absorbed or emitted. Now, that's for a single photon. If we have a lot of photons, generally the energy distribution of all the photons together is said to be described by some function, such as $I_\lambda$ (the subscript refers to wavelength). We can define the full pressure then by not only integrating over all wavelengths of all the photons, but importantly all the directions that the photons may hit or be emitted.

$$P = \frac{2}{c}\int_0^\infty\int_0^{2\pi}\int_0^{\pi/2} I_\lambda(\lambda) \cos^2(\theta)\sin(\theta)\:d\theta\:d\phi\:d\lambda$$

Note now that we've replaced $\langle S\rangle$ with $I_\lambda(\lambda)$. Also, since we're assuming this is pressure on a flat surface, we're not integrating over every angle, but only half of them $-$ any photon, in order to hit the solar sail, has to come from the direction the sail is facing, and can only reflect in the direction it came from, i.e., we're discluding photons "behind" the sail. For that reason, the $d\theta$ integral only goes to $\pi/2$.

If you don't understand the math at this point, that's fine but here's the important takeaway: If we only integrated over all wavelengths, you'd get the equation without the three that you listed. What that means is that the equation without the three doesn't take into account all the possible angles and orientations of the incoming photons. It simply assumes all your photons hit your surface perpendicularly, and then bounce off perpendicularly. This is of course not physically accurate as the photons can come in from a variety of angles.

If you do the angle integrals out, you find that you get the following:

$$P = \frac{2}{c}\int_0^{2\pi}\:d\phi \int_0^{\pi/2} \cos^2(\theta)\sin(\theta)\:d\theta\int_0^\infty I_\lambda(\lambda)\:d\lambda$$

$$P = \frac{2}{c}\frac{2\pi}{3}\int_0^\infty I_\lambda(\lambda)\:d\lambda$$

You can see there in the last line that the three popped out of those integrals over all the possible angles (specifically the $d\theta$ integral). Now, you get the rest of this by assuming some sort of function for $I_\lambda(\lambda)$ which describes the energy distribution of the photons as a function of wavelength. Since your light comes from a star, you know that's just going to be the Planck distribution (specifically the equation for $B_\lambda(\lambda,T)$). This integral is non-trivial, but if you again run through the math, you find

$$P = \frac{2}{c}\frac{2\pi}{3}\int_0^\infty \frac{2hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda k_BT}-1}\:d\lambda = \frac{2}{c}\frac{2\pi}{3}\frac{\sigma T^4}{\pi}$$

$$\boxed{P = \frac{4\sigma T^4}{3c}}$$

This is the true equation for radiation pressure on the solar sail, assuming reflection is occurring. This includes the three because we need to account for the photons hitting and bouncing off at all possible angles (for the solar sail). Note one important differences here is that my equation already has the extra "2" due to reflection factored in. Your last equation differs from mine in that it throws in that extra "2" for reflection. So either it erroneously included the factor of "2" twice, or (more likely) its account for reflection at every angle, not just all possible angles. Remember above, we only integrated the $d\theta$ angle up to $\pi/2$ for the reason explained there. If we integrated up to $\pi$ to include every angle, then we'd get an extra "2" as well and my equation would match your last one. Your solar sails though, don't receive or re-emit photons at every angle, and so my equation differs by that factor of "2".


Why would the Perseids meteor rate fall off after maximum faster than the increase before maximum? Writing about the 2018 occurrence of the annual Perseids meteor shower, the EarthSky.org Astronomy Essentials article Start watching for Perseids now says:

  People tend to focus on the peak mornings of the shower and that’s entirely appropriate. But meteors in annual showers – which come from streams of debris left behind in space by comets – typically last weeks, not days. Perseid meteors have been streaking across our skies since around July 17. We’ll see Perseids for 10 days or so after the peak mornings on August 11, 12 and 13. What’s more, the Perseids tend to build up gradually, yet fall off rapidly. So it’s possible to catch Perseid meteors now.

Looking at images in @userTLK's excellent answer to "What (actually) is Jupiter doing to this year's Perseids meteor shower?" one can imagine the Earth passing through a "fuzzy tube" around the orbit of the shower's associated comet's orbit, which is often inclined with respect to the ecliptic.

So far I can't imagine a mechanism that would make the ramp-up and then ramp-down happen at very different rates. Why would the Perseids meteor rate fall off after maximum faster than the increase before maximum?

I've just found this artist's conception of the "fuzzy tube" of debris in Sky and Telescopes's “Great Show” Predicted for Perseid Meteor Peak on August 12–13.

above: "Every year, in mid-August, Earth passes collides with particles spread along the orbit of Comet Swift-Tuttle." Credit Sky & Telescope, from here.

A very interesting question. Neither Bruce McClure at EarthSky nor I knew the answer, despite the fact that the observed gradual rise in numbers for the Perseids - and their quick drop-off - is something we've both observed for decades. So we asked a true expert on meteor showers, Robert Lunsford of the American Meteor Society. He also said: "Good question!" And, he said: "It seems that the particle density of the meteor stream [what you're calling here a 'fuzzy tube'] is greater prior to maximum than after. Perhaps the core of the stream, which produces maximum activity, is not quite centered in the stream. This is true for some other major showers such as the Geminids. The core of the Geminids must be really off-center as the display nearly disappears 48 hours after maximum. Gravitational effects from the sun may also play a role in this shift."

So here, as so often in astronomy, the answer suggests our knowledge of nature is - at present - inexact. Are any of you old enough to remember when the Voyager spacecraft first encountered Jupiter? Nowadays, we're used to seeing the very detailed images of Jupiter's cloud bands, but when we first saw that detail - via the Voyagers in the late 1970s - we were blown away! The situation with meteor streams in space is likely exactly the same. We picture them as symmetrical only because we don't have instruments refined enough to reveal their structure. All we have are the long-observed rates of meteors in annual showers - such as the Perseids and the Geminids - suggesting that meteor streams do indeed have a structure.

Hope this helps!


Unbelievably high masses/ densities in NASA website data In May, I did my own amateur analysis of Kepler data that was supplied here. This was a table of all confirmed "Kepler planets" to date (April 1, 2015). This table had some unbelievably high masses/densities listed in the first column after the planet designation (ie: planets of Kepler- 23, -24, -25, -27, -28, -32, -39, -48, -52, -54, -57, -58, -59, -60 and others). I didn't take these seriously, but would like to know if there was a "units" error or something I am missing. To give you an idea, Kepler-23b is given a radius of $1.9 R_{\text{Earth}}$ and a mass of $254.3 M_{\text{Earth}}$ - and that's nothing compared to some others. I checked just now and the data is the same.

Starting from the index you mentioned, I clicked through the links for some individual planets, which in turn link to discovery papers or other relevant observations. For planets around Kepler-23, -24, -25, -26, -27, and -28, the relevant papers are Ford et al. (2012) and Steffen et al. (2012), two out of a series of papers. Both papers used transit timing variations (TTV) and something called dynamical stability, which gives an upper limit to the mass of the planets such that the relevant n-body system is stable over long timescales. This leads to plots like the following (Fig. 3, Ford et al.):

The arrows indicate simulations where the system was stable over timescales greater than $\sim10^7$ years.

Additionally, as Ford et al. note,

  We caution that the uncertainty in the masses and sizes of the host stars directly translates into uncertainties in the planet masses and sizes.

Furthermore, shorter observations by Kepler also lead to larger uncertainties. Thus, the maximum masses of the planets as given in these simulations may very well be much greater than their actual masses. Steffen et al. have similar results, and they use a very similar method.

After more reading, it's clear that the same is true for planets around Kepler-32, Kepler-48, Kepler-52, Kepler-54, and Kepler -57 to -60, inclusive.

Kepler-39b is interesting. The paper NASA cites, Bouchy et al. (2011), states that the object is larger than one would expect, even taking into account that there is a large uncertainty in radius. It may be a fast rotator (see Zhu et al. (2014)), which would explain its oblateness, but that still isn't enough to explain the discrepancy.

Now, I referred to Kepler-39b as an "object", not a planet. NASA lists its mass as over 20 Jupiter masses; Bouchy et al. placed it at about 18 Jupiter masses, with a very low uncertainty. This could mean that it's more like a brown dwarf than a massive gas giant, and remember, the distinction is fuzzy when it comes to sub-brown dwarfs. This, then, may be the most likely explanation.


What is the minimum size of a ball of gas to become a star? I know there are two criteria to meet in order for nuclear fusion to occurs.

High temperature (many times temperature at Sun's core)
High pressure (protons are very close to each other)

[Goal]

However I want to know what equations allows me to calculate the amount of hydrogen gas needed so that the internal pressure can produce the right amount of heat for nuclear fusion to occur.

[My own understandings]

I also understand that the temperature of our Sun's core is not sufficient to overcome the coulomb barrier, it make use of quantum tunneling to achieve fusion instead.

[Question]

What is the minimum amount of hydrogen gas required to form a star? (also please include the name of the equation used)

[Assumptions]

shape is perfectly spherical
no angular velocity, not rotating
homogenous, consist of 95% hydrogen and 5% helium
sustain fusion for at least 1 million years

There is no equation, you need a detailed model of the interior physics of very low mass stars. Very roughly, you can say that
hydrogen fusion occurs when the central temperature exceeds about $10^{7}\ K$ (the density dependence is secondary) and that from the virial theorem, the central temperature is given approximately by $T \simeq 1.6\times 10^{7} M/R$, where the mass and radius are in solar units. $M/R$ decreases slowly down the main sequence towards lower mass stars, but then what happens is that electron degeneracy pressure becomes important and less massive objects do not become much smaller than Jupiter-sized and thus $M/R$ decreases and fusion stops - or rather it never starts.

In detail the minimum mass for hydrogen fusion in a manner that is capable of sustaining a star in equilibrium against gravitational contraction is about $0.075M_{\odot}$. With an uncertainty of about $0.002M_{\odot}$.

It is slightly more complicated than this, since at lower internal temperatures the deuterium in a star can fuse. This will happen during the early life of any ball of gas more massive than about 13 Jupiter masses.

Your assumptions about the relative H and He abundances are completely wrong. Even in the early universe the gas is $\sim 25$% He by mass.


How to Find the Viewing Size of a Star The sizes of the stars in the picture below have more to do with their brigtness than the actual size of the star.  I'm looking for a formula that relates the viewing size of a star with its brightness.

This is what I've worked on so far:

From Wikipedia on Apparent Magnitude, History

The viewing brightness can be related to the apparent magnitude by

$$B=0.4^A$$

Before I sit down and count pixels and try to do a curve fit to find the  the relationship, I'm guessing this work has been done before.  I wonder if anyone has any leads that would get me to a good equation that I can use to generate the view sizes of stars.

Edit:

My main goal is to make a reasonable 'cartoon' of the various constellations for the astronomy classes I'm teaching.  My rough work so far looks like this.  (I'm fairly certain the vertical scale is off.)

Based on the math in the Wikipedia article about the Airy Disk:

The intensity of the starlight as a function of angular distance $\theta$ from the center is reasonably approximated by a Gaussian $$ I(\theta) = I_0 \exp(-\theta^2/(2\sigma^2)) $$ where $\sigma$ is a measure for the width of the intensity profile, which for an ideal optical system and diffraction-limited images is $$ \sigma \approx 0.42 \lambda N $$ (where $\lambda$ is the wavelength and $N$ is the f-number) and which is independent of the peak intensity $I_0$.  For a non-ideal or not-diffraction-limited system, the Gaussian should still provide a reasonable approximation, but then $\sigma$ is (much) larger than for the ideal system.

The apparent edge of a star in an image occurs where the intensity $I(\theta)$ drops below a certain minimum $I_\text{limit}$, which (from the first equation) corresponds to an angular distance $$ \theta_\text{edge} = \sigma\sqrt{2\log(I_0/I_\text{limit})}$$

We can rewrite this in terms of visual magnitude $V$.  For fixed $\sigma$, $I_0$ is a measure for the brightness of the star.  Brightness is related to magnitude $V$ through $$ V = -2.5 \log_{10}(I/I_\text{ref}) $$ (with $I_\text{ref}$ the reference intensity for the magnitude scale) so $$\log(I_0/I_\text{limit}) = 0.4 \log(10) (V_\text{limit} - V_0)$$ and
\begin{eqnarray}
   \theta_\text{edge} & = & \sigma \sqrt{0.8 \log(10) (V_\text{limit} - V_0)} \approx 1.36 \sigma \sqrt{V_\text{limit} - V_0}
\\ V_\text{limit} - V_0 & = & \frac{\theta_\text{edge}^2}{0.8 \log(10) \sigma^2} \approx 0.543 \left( \frac{\theta_\text{edge}}{\sigma} \right)^2
\end{eqnarray}

So, the difference between the limiting magnitude and the star's magnitude is proportional to the apparent angular area of the star in the image.


Earth and moon gravitational ratios and proportionalities This question is for an astronomy class of mine. I don't believe the question is particularly difficult, but rather I'm having trouble understanding what it wants me to do. We haven't covered ratios and proportionalities and as a result I'm very lost. Here's the question:

The gravitational acceleration due to a planet or moon tells you how “heavy” you feel when you stand on the planet or moon.  Gravitational acceleration is proportional to the mass and radius of the planet or moon in the following way:
$g\propto \frac{M}{R^2}$

Start by writing two proportionalities, one for the Earth, one for the Moon. Use the following variables:

$M_{E}$ : mass of Earth
$M_{M}$ : mass of Moon
$R_{E}$ : radius of Earth
$R_{M}$ : radius of Moon
$g_{E}$ : gravitational acceleration of the Earth
$g_{M}$ : gravitational acceleration of the Moon

Following the example under the ratio section, divide the two proportionalities to form a ratio, i.e., the ratio between the gravitational acceleration of Earth and the gravitational acceleration of the Moon. In other words, find the following in equation form:
$$\frac{g_{E}}{g_{M}}$$

This isn't from a lack of effort. I'm just not entirely sure how to write or divide a proportionality.

The first part is just replacing $M$ with $M_E$ and so forth. You get two statements of proportionality, one with the $E$ values and one with the $M$ values.
The second part is algebraic division.  Mathematically if $a\propto  b$,  and if $a_1,a_2,b_1,b_2$ are specific values that are in this proportion. $a_1=kb_1$ and $a_2=kb_2$ for some constant $k$. Division gives $a_1/a_2 = b_1/b_2$.  The constants  cancel
By division, you turn a statement of proportionally into an equation.


What are the next planned space telescopes? We've had hubble for two decades. Its become the most famous space telescope, and lately Kepler seems to be running for the second place. Are there any more powerful, better telescopes planned for launch in the near future?

The James Webb Telescope is the next one on the launchpad that you might be familiar with.

Although there are a few differences that one ought to be aware of.  NASA has an entire program of telescopes to observe the universe, and many of them are designed for different wavelengths of light.  The James Webb is primarily designed for the infrared part of the spectrum.  While many people view the Webb as the successor to the Hubble, it would be more fair to say that it's really a successor to the Spitzer Telescope.  Hubble and Spitzer were both part of the "Great Observatories" program.  The Webb telescope is barely hanging on to funding, but so far the US hasn't managed to entirely abandon the program, science, or reality.


How to find if a set of objects lies within a given resolution (5 arcsec) on a telescope? I am new to astronomy and I was given the task of finding out if there are objects which can't be resolved by a telescope of resolution 5 arcseconds.

I was given the values of RA and Dec of these objects in degrees.

   RA              DEC
 201.1999388    27.49294258
 201.2048319    27.49212959
 201.1984702    27.4978016
 201.2025157    27.51610266
 201.1997827    27.51743447 ...etc

How can I find these in the simplest way? Is there any simple formula which can be used to achieve this which is computationally efficient as well?

This would be your best resource. Just note that the variables are expressed in geographical latitude and longitude, and your values are Right Ascension and Declination. Longitude and RA are the same, but Declination is 0 at the equator and is positive toward the north pole, whereas latitude is 0 at the north pole and is positive toward the south pole.

Don't forget to mind your units (it'll probably be in degrees since that's what your RA and Dec are). If you're doing this with a computer, you'll probably want to convert to radians first since most computer trig functions are in radians. Whatever program you use, just be sure to check what units it prefers.

Finally, be sure to convert your separation to arcseconds and you should be all set.


Did interstellar asteroid Oumuamua pass near other stars? "Oumuamua came from the direction of Vega, a bright star in the Lyra constellation. But even at 85,700 miles per hour, it took so long to reach our solar system that Vega wasn't in the same position 300,000 years ago."
CNN.

However, there is information on the motions of stars from ESA's Gaia mission (4 min. youtube with description), so it should be possible to project the motion of Oumuamua back it time to see if it was near any stars. Has anyone done this?

A recently released paper on arXiv (Bailer-Jones et al., AJ) has made use of the recently released GAIA-DR2 data set, and specifically the 7 million stars that have full 6D information, to try and find plausible home stars for 'Oumuamua. They found 4 potential systems of which the closest encounter was at 0.6 parsecs with the M2.5 dwarf HIP 3757 at a relative velocity of 24.7 km/s, 1 million years ago.  However all of the encounter stars have quite high encounter velocities.  A giant planet or a binary system could provide the needed velocity but none of the 4 systems are known to be binaries or to harbor planets. However 7 million stars with full information is still a pretty small fraction of the total, a situation which will improve with later data releases.


Where does the dust on 67P/Churyumov-Gerasimenko come from? As we (laymen) are told, comets lose gas and dust when they descent into the denser regions of the solar system. Seemingly so much and so large parts that they can even cause meteor showers like in the case of the Leonids from 55P/Temple-Tuttle.
Nevertheless we were also told, that the surface of 67P is covered with several centimeres of dust over a much harder surface. Where does the dust come from? The solar wind should have blown away any lose dust long ago, no? Is there a process which regenerates dust on the surface every cycle?

Your assumption that solar wind should have blown the dust away is only valid for comets that have a close perihelion. But you are correct in that there is a process which regenerates dust.
From Wikipedia

A comet will experience a range of diverse conditions as it traverses its orbit. For long period comets, most of the time it will be so far from the Sun that it will be too cold for evaporation of ices to occur. When it passes through the terrestrial planet region, evaporation will be rapid enough to blow away small grains, but the largest grains may resist entrainment and stay behind on the comet nucleus, beginning the formation of a dust layer. Near the Sun, the heating and evaporation rate will be so great, that no dust can be retained. Therefore, the thickness of dust layers covering the nuclei of a comet can indicate how closely and how often a comet's perihelion travels are to the Sun. If a comet has an accumulation of thick dust layers, it may have frequent perihelion passages that don't approach the Sun too closely.
A thick accumulation of dust layers might be a good description of all of the short period comets, as dust layers with thicknesses of order meters are thought to have accumulated on the surfaces of short period comet nuclei. The accumulation of dust layers over time would change the physical character of the short-period comet. A dust layer both inhibits the heating of the cometary ices by the Sun (the dust is impenetrable by sunlight and a poor conductor of heat), and slows the loss of gases from the nucleus below. A comet nucleus in an orbit typical of short period comets would quickly decrease its evaporation rate to the point that neither a coma or a tail would be detectable and might appear to astronomers as a low-albedo near-Earth asteroid.


What is that donut-shaped object I see in my telescope? I was looking through my telescope for the first time and came across this weird star. I think its a celestial donut but im not sure.

This is because your image is not in focus. So you’re seeing the shadow of the secondary mirror.
You should have a knob near the eyepiece, that you can turn to adjust focus. You need to turn it, one way or another, until the image is as small as possible and as bright as possible.
Good luck!


Does astro means space or stars In the popular TV show The Big Bang Theory Season 3, Episode 4, Sheldon Cooper said that astro means star and Raj said astro means space, I got a little confused here, various websites and others are going 50-50 on the subject. What does astro actually means?

Raj: No, no, no, no! That rate is much too low for what we’d expect from this collision. Do you understand that we’re talking about dark matter colliding in outer space?
Sheldon: Of course I understand. And who are you to tell me about outer space?
Raj: I’m the astrophysicist. Astro means space.
Sheldon: Astro means star.

Astro comes from the greek άστρο which not only means star but has a more general meaning that includes any small dot in the night sky (see https://en.wiktionary.org/wiki/%CE%AC%CF%83%CF%84%CF%81%CE%BF#Greek for details)

In today's language it still has a quite broad meaning:

Oxford English dictionary (https://www.oed.com/view/Entry/12240?isAdvanced=false&result=2&rskey=LlaQB9&):

  Forming terms relating to stars, other celestial objects, or outer space.

Merriam Webster (https://www.merriam-webster.com/dictionary/astr-)

  : star : heavens : outer space : astronomical

Conclusion: Sheldon and Raj are right


Can we, with arbitrary precision, calculate what's the place on earth that's closest to the sun on a given day? Can we, with arbitrary precision, calculate what's the place on earth that's closest to the sun on a given date? If so, what's the math like? Say, if I wanted to calculate this within 1m, 10cm or 1cm precision, what exactly should I take into account?

Calculation of the subsolar point is relatively straightforward. The actual point moves at nearly 500m/s in a circle, as the Earth spins. The linked website does the calculation in a few lines of python using a free library, The value is probably correct to within a few 10s of kilometres.

But you are looking for a different order of accuracy:  If you (for some reason) need 1cm accuracy, you would need to take into account the actual, non-spherical shape of the Earth.

The Earth is not a sphere, it can be better modelled as an oblate ellipsoid. But for 1cm accuracy, you would need to consider all of the mountains and hills. And you would need a very accurate time, not just a date. That is not a practical calculation. I can think of no practical application.


How much mass does the Sun lose as light, neutrinos, and solar wind? In this answer I estimate the loss of gravitational mass of the Sun to be about 1.3E+17 kilograms per year using $E = m c^2$ and 1360 W/m^2 at 1 AU.

While writing that answer I realized that solar neutrinos also carry away a lot of mass equivalence in the form of kinetic energy. The strongest component seems to be from the first step in the process; deuteron formation from proton–proton fusion.

Am I right so far?

Questions:

How much mass equivalent is lost per year by the sun via neutrinos?
Is my estimate of 1.3E+17 kg for loss per year via electromagnetic radiation close?
Are there any other comparable loss mechanisms? I'm assuming that the solar wind is lower, but I could be way wrong on that!

"bonus points:"

Are these rates likely to have been at least similar 4.5 billion years ago? Or would there have been large changes over this time? ("large" = factors of 2 or so)

The solar neutrino luminosity is about 2.3% of its electromagnetic luminosity (i.e. light). So the extra mass lost in the form of neutrino energy is 2.3% of your original calculation.

The average mass loss in the form of a wind and coronal mass ejections is about $4\times 10^{16}$ kg/year, but varies with the solar cycle (and from cycle to cycle) (Mishra et al. 2019).

4.5 billion years ago? It depends how exact you want to be. The Sun is thought to be 4.57 billion years old, so 4.5 billion years ago it would have been 70 million years old.

A 70 million year old Sun would have been on the hydrogen burning main sequence and about 20% less luminous than it is now, so you can scale your luminosity-mass and neutrino mass loss rates by about 0.8.

However, the solar wind was probably much stronger than it is now. The observational constraints on this are weak, but theoretical models suggest the mass loss rate in the wind scales as rotation rate $\Omega^{1.33}$ (Johnstone et al. 2015). Unfortunately, we still don't know how fast the Sun rotated in its infancy; it could have been anything from about 10 to 100 times it's rotation rate now. That means the mass loss rate in the wind would have been 20-500 times what it is now. Thus mass loss from a wind would dominate.

But maybe you meant $\sim 4.5$ billion years ago, in the sense that you wanted an answer for before the Sun became a star. i.e. Before hydrogen fusion began at a few million years after the Sun's birth. In that case, the wind losses might have been as per the 70 million year old case (with similar uncertainties), but there would be no neutrino losses (no nuclear reactions) and the luminosity of the Sun could have been a factor of 10 higher as a contracting pre main sequence star. In that case, mass loss from a wind would probably still be the biggest contributor.


Solar azimuth for celtic fire festivals I am pretty new to astronomy, and I have fallen down rabbit hole creating a new Metonic calendar in Javascript.

As part of this I would like to calculate when the sun is half way between a solstice and an equinox as viewed on a horison. Forgive me if my terminology is off but I believe this is known as the azimuth.

My issue is that I do not know what the correct values are. I am already  calculating the solstice and equinoxes using the suncalc javascript library

https://github.com/mourner/suncalc

in this code are the following 4 lines that determine the solstices and equinoxes. 0 is the spring equinox and 3 is the winter solstice.

switch( k ) {
     case 0: JDE0 = 2451623.80984 + 365242.37404*Y + 0.05169*POW2(Y) - 0.00411*POW3(Y) - 0.00057*POW4(Y); break;
     case 1: JDE0 = 2451716.56767 + 365241.62603*Y + 0.00325*POW2(Y) + 0.00888*POW3(Y) - 0.00030*POW4(Y); break;
     case 2: JDE0 = 2451810.21715 + 365242.01767*Y - 0.11575*POW2(Y) + 0.00337*POW3(Y) + 0.00078*POW4(Y); break;
     case 3: JDE0 = 2451900.05952 + 365242.74049*Y - 0.06223*POW2(Y) - 0.00823*POW3(Y) + 0.00032*POW4(Y); break;
   }

can anyone help me understand what the values might be to determine the halfway points between these solstice and equinox values?

thank you for your help.

The azimuth of the sunrise (or sunset, or any object) is a function of the Sun's declination and observer's latitude. It can be calculated from the following forumla:
$$\cos(\theta_R)=-\frac{\sin(declination)}{\cos(latitude)}$$
where $\theta_R$ is measured from due south to the location where the object rises or sets.

For example, at 55 degrees north latitude,

declination = 0 at equinox (by definition), so $\theta_R$ = 90 degrees from due south (90 degrees azimuth or due East when rising, 270 degrees or due west when setting).
declination = 23.4 (approximately) at the summer solstice, so $\theta_R$ = 133.8 degrees from due south (or an azimuth of 46.2 when rising, azimuth 313.8 when setting). [The declination at the solstice is plus or minus the obliquity of the ecliptic which is 23° 26′ 21″ = 23.4392° in the year 2000 and changes slowly.]
Half way between those two points on the horizon is (90+133.8)/2 = 111.9 degrees from due south. Solving the equation for the declination gives a value of declination = 12.4 degrees.
Now find the date when the Sun's declination is +12.4 degrees (April 23 and August 24 according to Table of the Sun's Declination).

Of course, the Sun will not be at the exact required declination at the time that it is rising (or setting), so you will want to select the date when the sunrise or sunset is closest. It will be accurate enough for most visual observations.


Calibrating raw photometric measurements So I have a photometry data sourced from Solar Mass Ejection Imager's (SMEI) server as an ‘add-on data’ for my point source of interest (a variable star) with a 104 minute cadence. However, it’s not really in the UBV photometry system, in V-band specifically, but rather scales with their SMEI photometry system while processed through a centroid fitting.
Is there a possible way to initially calibrate the photometry data like mapping the point source's nearby bright stars perhaps? And if so, how should I start of?
Attached photo is the 'raw' light curve of the star from the gathered SMEI data.

Please bear with me for I'm just a novice especially with regards to photometry. Guide(s) and response(s) are always welcome and I look forward to.
Clear skies.

If I understand the question correctly, you have raw data from a CCD sensor in arbitray units (which correlate with brightness) and your challenge is to callibrate this intensity for a certain frequency range.
Not knowing much about the SMEI sensor and where on the sky it has been looking at, I struggle to give you a full recipe. If you have for an object whose intensity (time series) from other observational instrument in the desired frequency band, you could use that non-SMEI data to correlate (if not even fit e.g. linearly) with the raw data time series for the same object as recorded by the SMEI sensor. This gives a calibration function which delivers intensity from raw data in arbitraty units.
References which sound possibly helpful

P. Hick, A. Buffington, B. V. Jackson: The SMEI real-time data pipeline: from raw CCD frames to photometrically accurate full-sky maps (2005)
A. Buffington et al.: Photometric Calibration for the Solar Mass Ejection Imager (SMEI)
Martin Clayton and Anthony Holloway: Simple Spectroscopy Reductions


Does Pluto have impact craters? I've heard from a lot of sources that Pluto doesn't seem to have any impact craters, but I see a few things that look like impact craters, highlighted in this picture:

For comparison, here's something I think looks like an impact crater on Pluto next to an impact crater on Mercury:

Are they impact craters on Pluto?  If not, what are they?  If they are, why are people saying it has no impact craters?

It was predicted by some, before the images arrived, that Pluto's craters would be much shallower than perhaps on other rocky objects/satellites. The reason is the icy composition and the relatively low impact velocities in the outer solar system (Bray & Schenk 2015).

There are indeed reported quotes in recent days that there are no impact craters on Pluto (e.g. here or here), based on the highest resolution images.

However, I'm with you - the things you have marked look like very large impact craters, or the remains of them, and so what we are seeing is that there are no small ones.

The gist of what I have heard/read is that this indicates geological activity on Pluto, which is effectively removing the evidence of relatively small impacts and that the surface we see is mostly younger than 100 million years old.
Watch this space, a lot of people are just flapping their lips at the moment (including me).

Another possibility - raised by Wayfaring Stranger - could be that the expected number of impacts was (hugely) overestimated. This is discussed by Singer & Stern (2015). There are indeed uncertainties, and it is hoped that cratering on Pluto & Charon would help constrain the number densities and size distribution of small impactors. My impression is that the uncertainties could be at levels of up to factors of 10 - and this seems to be discussed quite carefully in Greenstreet et al. (2015) (behind a paywall, sorry), who also comment in the abstract that surface ages judged from cratering are " entirely dependent on the extrapolation to small sizes [of the impactor distribution]".

Further Edit: LDC3 suggests all the craters could be too small to be seen. The high resolution images that have been released could resolve craters of a diameter of a couple of km or more and cover a few percent of the surface. According to Singer & Stern (2015), the number of craters at this size and above on Pluto they expected in such an area is $\sim 1000$. So I think that either the planet has been resurfaced recently, the craters have been eroded in some other way or their estimates for the number of impacts/size of the impactors is wrong.


How does science knows what is the early universe? I've read about Hubble seeing the brightest quasar in the early universe. Question is, how does science knows what is early and what is late?

Please correct me if I'm wondering, but after the big bang, everything went in all directions equally. How do we know what direction is the center of the big bang? And continuing on the line of thought, is "early universe" considered closer to the origin of the big bang or further away?

I assume you're referring to the recent press release about the quasar J043947.08+163415.7, observed recently using Hubble. The paper about the observations details how the authors measured the distance to the quasar, by calculating its redshift, a quantity that describes how the wavelength of light appears to change based on whether the object is moving relative to an observer. On cosmological scales, redshift can then be converted to distances. We usually see quasars at $z>0.1$.

As is usually the case, the quasar's redshift was determined by looking at spectral lines - in this case, Mg II emission around 21000 angstroms (see in particular the inset):

This emission line was observed to have a different wavelength than it would if the quasar was at rest, enabling the astronomers to calculate its redshift: $z=6.511\pm0.003$.

This is the typical process used to determine redshift, and therefore distance - well, not necessarily using that particular Mg II line, of course, but nonetheless the use of spectroscopy to measure how the location of spectral lines changes. The difference between "early" and "late" isn't quite clear, and probably varies depending on who you're talking to. For instance, someone studying the Big Bang might refer to the period before recombination as "early" - although we're seeing the quasar as it existed much later than that; it wasn't around for recombination.

I'd like to note that it's somewhat misleading to call this quasar the brightest (or even brightest known) quasar in the universe, because - as that press release notes - the gravitational lensing by a foreground galaxy, which enabled the object to be discovered, also magnified its brightness. It appears to have a brightness of 600 trillion solar luminosities, but without the lensing, that number is reduced to 11 trillion solar luminosities - a much more modest figure.


Gravitational lensing in Newtonian physics Famously, when Eddington attempted to measure gravitational lensing during the Eclipse, it was the measured magnitude of the lensing that gave gravity [pun, obviously, intended] to General Relativity - not the measurement of lensing itself. That is, Newtonian physics also predicted a lensing deflection, but only half of the deflection predicted by GR.

Question is: Why? I've read a lot about this, and I can see how when one integrates the Newtonian acceleration along the photon path with constant |v|=c this gives the 'correct' Newtonian value - but intuitively, I can't wrap my head around it. Why should a massless photon be affected by gravity at all in Newtonian gravity?

A photon is an entity defined in the context of a relativistic field theory, and so it doesn't really make sense to talk about the Newtonian bending of a photon. Necessarily, we need to substitute an analogous question that's sensible in the Newtonian framework. To do so, we can imagine a classical corpuscle of light--appropriately enough, a theory of light advanced by Newton himself.

There are many problems with the Newtonian conception of light, but that's an issue for electromagnetism rather than gravity.

Critically, for a test particle, the trajectory depends on the initial velocity only, and not the mass. So we don't even have to address the mass of the light corpuscle at all to talk about its trajectory, as the acceleration is the gradient of the gravitational potential and if considering force (or gravitational potential energy), the mass cancels out anyway. If you wish to bring in the mass explicitly, we can still think of the trajectory of a massless particle as a limit of trajectories of particles with masses tending to zero but having the same velocity — a trivial limit because they're all the same trajectory in the Newtonian theory. On the other hand, if we recognize that light carries momentum, a Newtonian light corpuscle shouldn't have zero mass, so the question of what to do with a genuinely massless particle evaporates.

When people talk about the Newtonian deflection of light, they are typically considering a hyperbolic trajectory of test particle at the speed of light under Newtonian gravity. If the angle between the asymptotes is $\theta$, then $\theta = \pi$ represents a completely straight trajectory unaffected by gravity, and the eccentricity is
$$e = 1 + \frac{v_\infty^2R_p}{GM} = \frac{1}{\cos\left(\frac{\theta}{2}\right)} = \frac{1}{\sin\psi}\text{,}$$
where $R_p$ is the periapsis distance and $2\psi = \pi-\theta$ is the measure of deflection. On the scale of the solar system, it doesn't matter whether we set $v_\infty = c$ or anywhere else along the trajectory. For example, if the velocity at peripasis is $c$ instead, then $e\mapsto e-2$, and so is negligible for light deflection due to the Sun, $e > 10^{5}$. The total deflection is approximately
$$2\psi \approx e^{-1} \approx \frac{2GM}{c^2R_p}\text{,}$$
which is half the correct general-relativistic prediction.

Note that here we did not assume that the speed is constant along the hyperbolic orbit. That would not be consistent with Newtonian gravity. Rather, what we have is a situation where if $v = c$ anywhere along the trajectory, then the speed along any other point is so close to $c$ that it doesn't practically matter for considering light deflection.


Why is a meteorite impact on the moon visible as white light? You can watch a video of a meteorite impacting the moon at the npr.org website.

Why does the impact produce white light?

I understand that the impact is like an explosion, but there is no air on the moon to burn material. If it was the result of superheating rock from the impact, then why does it fade so fast. White light emitted by heated rocks should last longer as it takes time to cool.

We can see meteorites breaking up in Earth's atmosphere as a streak of light, but there is no atmosphere on the moon.

So why isn't it just a big black puff of smoke?

The velocity of the impactor of tens of kilometers per second provides enough energy to heat the impactor and parts of the target to several thousands of Kelvins, so that parts are converted to plasma or to vapor, at least.

According to Planck's law the color at these temperatures is white or bluish.

According to the Stefan Boltzman law the total emitted energy is proportional to the forth power of the absolute temperature. Hence material which isn't heated to thousands of degrees, and may glow yellowish or reddish is too dim to be noticible in comparison.

Since most of the heated ejecta are finely distributed, they cool down rapidly by the emitted radiation, and also after contact with the surface.

After cooling there will certainly be some fine dust, which still has to fall back to Moon's surface, but you won't see this on a video from a distance, since it's cooled down and distributed over a wider area.


What is the magnification of my new telescope? I just bought a Skywatcher telescope with a diameter of 60 mm and a focal length of 900 mm. I have a 2x Barlow lens and two eyepieces with focal lengths of 10 mm and 20 mm.
How can I calculate the magnification of my telescope, for example using the 2x Barlow and the 20mm eyepiece?
I'm fascinated by watching the sky and I can't wait to try it... Unfortunately, it's currently cloudy.
One more question; what I can see to be fascinated with this telescope?

Magnification of your telescope depends on the ratio of the focal length of your primary optics and the focal length of your objective. We can represent this with this simple formula:
$$P=\frac{f_{objective}}{f_{eyepiece}}$$
where $P$ is magnification (power) and $f$ is focal length.
Your telescope has 900 mm focal length, so the magnification is $P_1=\frac{900\text{ } mm}{10\text{ }mm}= 90\text{x}$ (10 mm ocular) and $P_2=\frac{900\text{ }mm}{20\text{ }mm}= 45\text{x}$ (20 mm ocular).
A Barlow lens is used to increase the focal length by the given factor (thus 2 in this case), so directly impacts the magnification in the same way, increasing it to 180x and 90x - but gives you in return an equally smaller field-of-view and makes area-objects fainter as the same light is projected to a larger area. Note that there is a maximum useful magnification which is roughly twice the aperture (measured in mm), which is 120x in your case, so the Barlow lens will not be very helpful as it magnifies beyond reasonable due to inevitable diffraction on the clear aperture.
One usually compares them by their focal length and by their aperture - the ratio defines how bright the objects appear and thus how faint objects you can see.
With an aperture of 60 mm, it has an opening ratio of f/15 which is only a moderate one.
Such telescope is particularly useful for observations of the planets in our solar system as well as star clusters and the brighter deep-sky objects. Now in summer would be an excellent time to hunt for the globular clusters like the one in Hercules (M13). Also, you can download a useful program Stellarium to find more interesting features.


Can Drizzle Integration increase resolution beyond theoretical resolution of telescope? So I was wondering, if drizzle integration actually increases details in your image if the drizzled image results in higher resolution in terms of arcseconds / pixel than the telescope would allow (due to the uncertainty principle the telescope's resolving capability is dependent on its diameter, more commonly referred to as the diffraction limit 1, 2 )

In my case I have around 0.52" per px on my main camera and my GSO RC8" allows for around 0.5" resolution max due to the uncertainty principle. Drizzling an image would then give me 0.26" per px on the drizzled image when increasing width and height of the image by factor 2, which is higher resolving power than what the telescope would actually be able to do.

Apart from sharper and more defined stars when drizzle integrating, will drizzle integration actually increase my resolving capability?

Drizzling can't actually do any better than the theoretical resolving power of the combination of your telescope and the atmosphere. What it can do is at least partly compensate for having pixels that are too large to properly sample the resolution of your telescope + the atmosphere.

The Nyquist Sampling Theorem basically says that you should have at least 2 pixels per resolution element in order to properly sample the resolution. Or, conversely, if you want to achieve a resolution of X arcsec, you want pixels that are no more than X/2 arcsec in size. So if the theoretical combination of your telescope diameter and the atmosphere had a resolution of 0.5 arcsec (so we're talking about really good atmospheric conditions), you want pixels that are 0.25 arcsec or smaller; otherwise, the pixelization overwhelms the intrinsic resolution. But note that you don't really gain anything from going to even smaller pixels: you'll just have a nice blurry image without visible pixelization. Having pixels that are 0.1 arcsec in size won't make the image any sharper than pixels that are 0.2 arcsec in size, if the resolution of your system (telescope + atmosphere) is ~ 0.5 arcsec.

In your hypothetical case, 0.52-arcsec pixels are good enough for resolutions down to about 1 arcsec. Ideal drizzling would allow you to effectively have pixels that were about 0.25 arcsec in size, which would accommodate a resolution of about 0.5 arcsec. But you can never do better than the intrinsic resolution of your telescope + the atmosphere.


Birth of the Gas Giant Planets? How does a gas giant planet form like Saturn and Jupiter, and why just gas? Why didn't they form any solid surface?

As @RobJeffries says, this information is mostly in the relevant wikipedia article.

The answer is really in two parts:

Firstly, they don't have solid surfaces because the elements they are made of don't have solid forms at the temperatures and pressures that arise inside the planets. In particular, Jupiter and Saturn are made mostly of hydrogen and helium which only become solid at rather low temperatures and the planets are still hot due to the heat created by their formation, heating from the Sun and some heat released by internal processes.

This, of course, begs the question of why those two planets are made mostly of hydrogen and helium, when the others are not. I should start with the caveat that a lot of aspects of this are still areas of active research and debate among scientists, and could not be considered settled  or universally accepted.  As more evidence is gathered and computer models (used for making checkable predictions from underlying theories) are refined,views of this may change.

So the solar system started out as a cloud of gas and dust. While it was mostly hydrogen and helium (still around since the big bang) it had significant amounts of other elements present in it which had been produced in older generations of stars. Some chance concentration of mass, or external shock (eg from a supernova) caused it to start to collapse and, under the influence of its own gravity, fluid dynamics and the conservation of angular momentum, a large part of it formed a ball (called a protostar) at the centre which was heated up by the collapse process itself. Around that, gas and dust formed a flat disk (oriented according to the original net angular momentum of the clump). Within that disk, tiny grains of solid matter formed. Far from the protostar, those grains were mostly made of compounds relatively light elements (carbon, oxygen, nitrogen, hydrogen) which we collectively call "ices". Closer to the protostar, its heat stopped ices from solidifying, but still allowed grains of iron, rock-like materials and so on.

When these grains collided, they stuck together, gradually forming bigger and bigger lumps. In the outer system they were lumps of ice, a lot like comets, in the inner system, something more like lumps of rock (like some asteroids). These then started holding together and pulling in more material by their own gravity and eventually grew into planets. At some point during this process fusion started in the Sun, and drove awat most of the remaining gas and dust. After that the planets grew by collisions.

So in the inner system, you got rocky planets like Mercury, Venus, Earth and Mars. A bit further out you got icy planets like Uranus, Neptune and Pluto. A couple of those icy planets  happened to get big enough, quickly enough, to attract a lot of hydrogen and helium gas before the Sun lit up. They are Jupiter and Saturn.


A comet orbits the Sun in the same plane as Earth's orbit If the Earth around the Sun follows a practically circular orbit of radius equal to one astronomical unit, determine its orbital speed in astronomical units per year.
If the Earth around the Sun follows a practically circular orbit of radius equal to one astronomical unit, determine its orbital speed in astronomical units per year.
A comet orbits the Sun in the same plane as Earth's orbit. The comet's closest approach to the center of the Sun is 0.5 u.a. and it has, at that point, an orbital velocity of 4pi u.a / year. What is the orientation of the comet at the moment it cuts the Earth's orbit

I'll just give an outline about how I might try to answer these questions, rather than answering them outright.
Answer the first question using the circular form of the Vis-Viva equation: $v=\sqrt{\frac{\mu}{r}}$, where $\mu$ is the standard gravitational parameter ($4\pi^2 AU^3/yr^2$) and $r$ is 1 AU.  If your answer isn't $2\pi$ AU per year, you probably did something wrong with the units.
For the second question, try to determine the shape of the comet's orbit using the inequality for escape velocity: $v_e \ge \sqrt{\frac{2\mu}{r}}$.  Once you have the orbit shape, the rest of the problem falls out pretty easily using known equations for that orbit.

 Escape velocity at 0.5 AU is
 $$v_e=\sqrt{\frac{2*4\pi^2}{0.5}}=\sqrt{16\pi^2}=4\pi$$

 which is the exact speed given in the question. That means the orbit
 is parabolic.  From  Bogan's table pulled from
 Bate/Mueller/White's Fundamentals of Astrodynamics, for a parabolic
 orbit, $$r=\frac{2q}{1+\cos{\theta}}$$ where $r$ is the distance from
 the central body, $q$ is the closest distance of the orbit, and
 $\theta$ is the angle from periapsis or true anomaly. Solving gives us
 $\theta=\pi/2$.

 The same table gives us the angle of velocity relative to the
 perpendicular to the radial direction: $$\phi = \theta/2 = \pi/4$$

 So the orientation of the comet's velocity is at
 $\pi/4$ or 45 degrees off a tangent line to a circle around the Sun
 with radius 1 AU as the comet passes through it.


What is the exact mass of the Sun? I am having trouble finding data on this. So far, I have found at least seven values:

1: 1.9885  e30 kg
2: 1.98855 e30 kg
3: 1.9891  e30 kg
4: 1.991   e30 kg
5: 1.989   e30 kg
6: 1.99    e30 kg
7: 2       e30 kg

The first value can be found from NASA GSFC.
The second value can be found from Wikipedia in (at least) two places.

The first Wikipedia occurence references the GSFC page--which is erroneous since that value does not occur there.
The second Wikipedia occurence references USNO's 2014 Astronomical Constants, which appear to themselves be copy-pasted from data in 2009 or 2012.  The second Wikipedia occurrence also sources NIST (presumably this table).  I couldn't find the cited mass from either reference.

The third (and also the fifth) value (which, it is interesting to note, are both outside the error bound given in the second Wikipedia occurrence) can be found on this generic NASA site.
The fourth value occurs in the book Handbook of Chemistry and Physics (Robert C. Weast, 1980)
The fifth value apparently also occurs in the book Astrophysical Data (Kenneth R. Lang, 1990ish).
The sixth value occurs in the book Physics--3rd Edition (Cutnell et al. 1995).  This site lists several of the values and does a calculation to obtain the sixth value.
Occurs in many places and is pretty clearly rounded.

The fifth, sixth, and seventh values (also) occur in various non-primary sources online, and I assume are rounded values of others.

My question: what is the most recent, best estimate, to the greatest precision, of the Sun's mass?  Obviously, citable, primary sources only!

The mass of the Sun is determined from Kepler's laws:

$$\frac{4\pi^2\times(1\,\mathrm{AU})^3}{G\times(1\,\mathrm{year})^2}$$

Each term in this component contributes to both the value of the solar mass and our uncertainty. First, we know to very good precision that the (sidereal) year is 365.256363004 days. We have also defined the astronomical unit (AU) to be 149597870700 m. Strictly speaking, the semi-major axis of the Earth's orbit is slightly different, but by very little in the grand scheme of things (see below).

At this point, we can solve for the product $GM$, known as the gravitational parameter, sometimes denoted $\mu$. For the Sun,
$$\mu_\odot=132712440018\pm9\,\mathrm{km}^3\cdot\mathrm{s}^{-2}$$

So solve for $M_\odot$, we need the gravitational constant $G$, which, as it turns out, is by far the largest contributor to the uncertainty in the solar mass. The current CODATA value is $6.67384\pm0.00080\times10^{-11}\,\mathrm{N\cdot m}^2\cdot\mathrm{kg}^{-2}$, which all combines to give
$$M_\odot=1.98855\pm0.00024\times10^{30}\,\mathrm{kg}$$
where my uncertainty is purely from the gravitational constant.

The value $1.9891\times10^{30}\,\mathrm{kg}$ (and nearby values) probably come from an older value of the gravitational constant of $6.672\times10^{-11}\,\mathrm{N\cdot m}^2\cdot\mathrm{kg}^{-2}$, which is still supported by some measurements of $G$.


Do more objects appear on sky in time? I would assume that, with light from more distant areas in space reaching the earth, we should be able to see more distant (and older) objects.
However there's lot of weird stuff regarding the space expansion that could eventually make the objects leaving us so fast the we'd never see them.

So, did we reach any boundary of what we can possibly see, or are we still working on more sensitive exploration devices?

I'll be grateful for some useful re-tag

No, the expansion of the universe appears to be accelerating. Hence distant objects get more and more red-shifted. Our horizon is shrinking instead of expanding. At the end, only galaxies which are sufficiently bound by gravity remain visible; this could turn out to be the Virgo supercluster.

More reading about the accelerating universe on Wikipedia, and on National Geographic.

Scientific background to the according Nobel Prize in physics 2011.


Sun's formation as "part of a star-forming-cluster..."? After reading this interesting answer,

I was wondering, do we in fact know if our Sun in particular was created as part of a star-forming-cluster, or, was it more of a solo creation?

(Or, are all stars created as part of star-forming-clusters, is that the norm and how all stars come to be?)

If yes, indeed are we aware of which stars around us, are our sibling star?  (So, there's a list somewhere "stars known to likely be sibling to our Sun from a certain star-forming-cluster".)  Or have they dispersed over the 4 billion years and it's impossible to know, or is it just the case that "all the stars anywhere nearby" are/were indeed siblings from "our" star-forming-cluster?

Further, I'm having trouble grasping the order of magnitude of star-forming-clusters. Consider say "all the stars in our arm of the galaxy": Consider the question "in how many star-forming-clusters did all those stars originate?"  Is the answer like "3", "a few hundred" or "100 million" ??

Perhaps most simply: if sol formed as part of a star-forming-cluster, what's the order of magnitude of how many stars formed with it?

Fortunately, there is an extensive and highly respected Annual Reviews of Astronomy and Astrophysics article on this very question - Adams (2010), https://arxiv.org/abs/1001.5444

At the moment it is debated whether most stars form in clusters. They may do, but there is a lot of evidence for hierarchical structure formation that results in lots of stars forming in relative isolation.

Regardless, there are a number of strands of evidence that can be used to suggest that the Sun formed in a cluster of somewhere between a thousand and ten thousand siblings - a bit like the Orion Nebula cluster.

Adams argues for this on the basis of: (1) We have a relatively well-behaved solar system inside the orbit of Neptune. This argues the Sun did not (or rather was not likely to) form in a massive and dense cluster where it suffered lots of very close encounters. On the other hand, the wacky orbits of some more distant objects like Sedna do suggest some early close encounter with another star was required. (2) If the Sun was born in a very massive cluster, it would likely have been born close to some very massive stars. The UV emission from such stars could photoevaporate a circumsolar disk before it formed planets. (3) In tension with this is the probability that the protosolar disk was enriched with radioactive supernova products. This requires the Sun to be born close to moderately massive stars (that would explode on a short timescale), which are only found in moderately massive clusters.

Adams puts these constraints together to estimate a probability that the Sun was born in a cluster of a particular size. Figure 7 of the review shows this to be reasonably sharply peaked between a thousand and ten thousand.

An order of magnitude uncertainty is at least warranted and some of the assumptions and calculations are open to criticism. But it is a starting point for any further research into the matter.


When was the concept of rogue planets first theorized? So after watching the Kurzgesagt video about life on rogue planets I became fascinated by the concept. I did some digging, and the earliest reference to the concept I could find was Philip Wylie's When Worlds Collide: a Sci Fi novel written in 1933. In the novel Wylie writes

  A sort of body that they knew existed by the millions, probably, all through the universe-something they were sure must be, but the general existence of which has never been actually proved.

The best I could come up is with after LePlace's nebular formation hypothesis became more widespread in the 19th century people just assumed rogue planets would exist. I certainly can't find anything written about them before Wylie's novel. It doesn't help that the first use of the term 'rogue planet' I could find was George R. R. Martin's novel Dying of the Light and any earlier mentions than that use several different terms to illustrate the concept.

tl;dr Who first theorized the concept of rogue planets and when did they do so?

I don't know the first use of the time rogue planet or the first use of such planets called something else.
But if science fiction is anything to go by, rogue planets are as thick as fleas in interstellar space and they are all heading for Earth's solar system.
The Wanderer (1965) by Friz Leiber involves a planet entering the solar system and causing devastation on Earth.
The Fifth Planet  (1963) by Sir Fred Hoyle and Geoffrey Hoyle involves another star passing close to ours, and an expedition to a planet of that star.
Battle of the Worlds (1961) is a movie in which a rogue planet called "The Outsider" enters our solar system heading toward Earth.
Back in the 1960s I read a juvenile science fiction novel in which a planet with human like inhabitants entered the solar system.  I don't remember the title or author, but it is in the future when Earth has space travel, and strange spaceships are seen. The boy protagonist becomes friends with a boy from the alien planet.  I remember a scene where an Earth spaceship was struck by an energy beam hear its needle nose, and incandescent vapor sprayed out from the ship.  I think that the aliens looked like humans with orange skin.  This book could be the Winston science fiction The Mysterious Planet, 1953, by Kennith Wright (Lester Del Rey).
In The Secret of the Ninth Planet, 1959, by Donald A. Wollheim, it is discovered that there are Plutonians and they are up to no good.  It is discovered that Pluto was lost from its original star system and wound up in our solar system.
In The Mysterious Planet, 1953, by Kenneth Wright (lester Del Ray), a new planet, Planet X, is entering the solar system and there are mysterious attacks by space pirates on Earth spaceships. This is a Winston juvenile science fiction novel, and possibly the one I mention two paragraphs above.
When Worlds Collide  (1951) is a movie based on the novel, Two rogue planets named Bellus and Zyra are entering the solar system and Bellus will collide with Earth.
The Man From Planet X (1951) has an alien land in a small village in Scotland. Apparently he is from a rogue planet passing through our solar system.
In Chapter One, "The Planet of Peril", of the movie serial Flash Gordon (1936) The planet Mongo is headed on a collision course with Earth.

The planet Mongo is on a collision course with Earth. Dr. Alexis Zarkov takes off in a rocket ship to Mongo with Flash Gordon and Dale Arden as his assistants. They find that the planet is ruled by the cruel Emperor Ming, who lusts after Dale and sends Flash to fight in the arena. Ming's daughter, Princess Aura, tries to spare Flash's life.

https://en.wikipedia.org/wiki/Flash_Gordon_(serial)
As I remember, Ming says he controls the course of Mongo.  When Ming says he plans to destroy Earth Dr. Zarkov asks him why destroy Earth, why not conquer it, and Ming agrees.  So presumably Ming changes the course of Mongo and the planets don't collide.
The Flash Gordon Sunday newspaper comic strip ran from 1934 to 2003, and a weekday Flash Gordon comic strip ran from 1941-44, and 1951-1993.

The comic strip follows the adventures of Flash Gordon, a handsome polo player and Yale University graduate, and his companions Dale Arden and Dr. Hans Zarkov. The story begins with Earth threatened by a collision with the planet Mongo. Dr. Zarkov invents a rocket ship to fly into space in an attempt to stop the disaster. Half mad, he kidnaps Flash and Dale and they travel to the planet. Landing on the planet, and halting the collision, they come into conflict with Ming the Merciless, Mongo's evil ruler.[1][3][9]

https://en.wikipedia.org/wiki/Flash_Gordon#Comic_strip_characters_and_story
When Worlds Collide  (1933) is a novel by Philip Wylie and Edwin Balmer, in which planets Bronson Alpha and Bronson Beta enter the solar system, with Bronson Alpha on a collision course with Earth.
In "Wandl the invader" Astounding Stories (1932, book  version 1961), by Ray Cummings, the tiny planet Wandl is sent from another star system to invade our solar system.
A comment mentioned rogue planets in works of Olaf Stapledon.  Possibly one or more might be in Last and First Men (1930).
In "A Brand New World" Argosy All-Story Weekly (1928, book version 1964) by Ray Cummings, a new planet with humanoid inhabitants wanders into our solar system. If I remember correctly, the new planet is named Xeneprene.
In "The Star" by H.G. Wells (1864), a star enters the solar system on a collision course with the Sun and causes great damage when it passes close to Earth.

This story is often credited with having created a science fiction subgenre depicting the impact event of a planet or star colliding, or near-colliding with Earth—such as the 1933 novel When Worlds Collide by Philip Wylie and Edwin Balmer (made into a film in 1951), Fritz Leiber's The Wanderer (1965), and Lucifer's Hammer by Larry Niven and Jerry Pournelle (1977).
However, it was preceded by two stories in 1894: Omega: The Last Days of the World by Camille Flammarion (the astronomer of the Flammarion Catalog) and Olga Romanoff or, The Syren of the Skies by George Griffith. In 1895, Griffith used an comet disaster again in The Outlaws of the Air.

https://en.wikipedia.org/wiki/The_Star_(Wells_short_story)
I note that the comets that strike Earth in the Flammarion and Griffith stories are not necessarily described as from beyond the solar system.  By the 1890s it was probably realized that the vast majority of comets are solar system objects in orbit around the Sun.
I believe that Edgar Allen Poe wrote a sort story in which humans became extinct or Earth was destroyed, possibly by collision with another astronomical body.  This may be "The Conversation of Eiros and Charmion" 1839, in which two spirits of dead persons discuss the past destruction of the Earth by a comet. I don't know if the comet was supposed to be a solar system comet or a rogue comet.
And those are all the examples I could remember or dig up.
In real life interstellar objects should often pass through the solar system, most of them tiny asteroids and comets ejected from their star systems.  A number of candidates for those objects have been suggested, and there are two confirmed cases: Oumuamua, an extra solar asteroid in 2017, and 21/Borisov, an interstellar comet in 2019.
https://en.wikipedia.org/wiki/Interstellar_object#Confirmed_objects


Could a closely orbiting pulsar be used to study a black hole? If we were to locate a pulsar in a close elliptical orbit around a black hole, could this reveal significant information about the black hole (e.g. time dilation / structure)?  Could this even tell us the structure of a black hole (e.g. if it is a singularity, or a sphere formed around the event horizon, or something else)?

Yes. And we are actively searching for these systems, e.g., with the Pulsar timing array and "soon" with instruments like SKA.

From the Astro2020 Science White Paper on Fundamental Physics with Radio Millisecond Pulsars:

  A broad class of alternative theories invoke mediation of gravity through both tensor and scalar
  fields, whereas general relativity relies strictly on a tensor-field description. A key prediction from
  tensor-scalar theories is dipolar gravitational-wave radiation in compact binary systems with large
  differences in component binding energies (Eardley, 1975).

And:

  Future improvements to tensor-scalar tests will also come in the discovery of pulsars tightly
  orbiting stellar-mass black holes. Shao & Li (2018) recently estimated that a small but detectable
  pulsar/black-hole binary population – between 3 to 80 such systems – resides within the Galactic.
  disk

And about super-massive black holes:

  A key science goal for future radio-astronomical observatories is the discovery and timing of radio
  pulsars in orbit around the supermassive black hole residing in the center of the Milky Way galaxy
  (e.g. Bower et al., 2018). Recent projections have shown that an entirely new class of tests can
  be achieved with pulsar/black-hole orbits shorter than 1 yr in period, even if only one pulsar is
  discovered and yields low timing precision (Liu et al., 2012). These tests will directly probe mass
  and spin properties of the Galactic-centre black hole, as well as the validity of the storied “nohair” theorem (e.g. Will, 2008) with measurements of mass, spin and the quadrupole term of its
  gravitational potential.

So, you see, these source will be great for testing GR in general and studying the structure of BHs (or whether there is any) in particular.

Btw, such gravitational time delays, are often called Shapiro delay .


Why is the eccentricity of Triton so low? Triton is very special, since its orbit is retrograde respect Neptune, its obliquity is near 0° and its orbit is almost circular (according to Harris (1984) is $5 \times 10^{-4}$). I'm familiar with the theory that claims that Triton was several Myr ago a KBO that was captured by the gravity of Neptune, but I didn't find an actual explanation of why the orbit is almost circular. Can anybody give me some theories that actually explain the cause of this? I've heard about tidal circularization, but I don't know if that explains it at all...

Tidal circularization is indeed a plausible mechanism for why Triton's orbital eccentricity is so low, and I believe it is in fact the prevailing theory, at the moment. In short, tidal forces from Neptune dissipate energy in Triton's interior, essentially "squeezing" the moon. This results in a loss of orbital kinetic energy. The exact calculations that lead to the decrease in eccentricity are a little complicated - and described in the paper you linked to - but in a nutshell, Triton's orbit could have circularized on a timescale of about $\sim10^7$ to $\sim10^8$ years (Ross & Schubert (1990), for instance, computed the latter order-of-magnitude timescale).

Tidal circularization is not the only possible explanation for Triton's low eccentricity. Cuk & Gladman (2005) hypothesized that there was once a debris disk around Neptune, possibly caused by the collision of two or more other satellites after gravitational interactions with Triton. Interactions with the disk then contributed to energy loss from the moon in the same way that tidal circularization might have, thus leading to a smaller eccentricity. Interestingly enough, though, this could have happened within only $\sim10^5$ years! The authors have suggested that a hybrid of the two models may be the best explanation.