If you also have photometry, then there is a reasonably accurate conversion between the apparent magnitude and the continuum flux (per unit wavelength interval) at the wavelength of the photometry.
Once you have this conversion factor, you multiply your equivalent width by it to get a flux.
Visibility of a celestial body on the sky during the day depends on a number of factors - if I recall correctly: the body's magnitude, Seeing, Sky brightness, and probably some others I have missed.
Who was the first to discover/compile the set of variables responsible for determining, whether a celestial body remains visible with naked eye in certain (daylight) conditions?
I'm slowly starting to interest in astronomy. Currently I'm enjoying in stargazing but unfortunately place where I live is very light polluted.
I'm planing my vacation in a couple of months and one of the attractive places on my list is Tenerife because I can combine both star-gazing and going to the beaches. As per my research, base of Mountain El Teide is perfect for star-gazing so I tried to simulate El Teide location and date in the Stellarium software to see what I can expect. Based on the light pollution index (2) which Stellarium picked for me based on the location I got following picture:
My question is: How this rendered picture will match my naked eye in the reality at the same location and time (we can assume that there will not be any clouds and that I'll be stargazing when moon is couple of days from New Moon phase). Can I expect something like this or even something better?
I'm aware of all the tips which can maximize my experience but I'm trying to just be sure that El Teide will not disappoint me. I really want have great star gazing-experience and try to see Milky way and much stars as possible because I don't have that luxury in the country where I live.
Any personal star gazing experience from Tenerife location is more than welcome. For my first time I'll plan to not use any equipment (except maybe binoculars)
Btw, there are stellarium settings used for picture that I attached (I assume that this is closest what you can see with naked eye).
2010 TK7 has a diameter of about 300 meters (1,000 ft). Its movements do not bring it any closer to Earth than 20 million kilometers (12.4 million miles), which is more than 50 times the distance to the Moon. (from the wiki entry for it)
That should make it pretty clear? Can you see a fly from 200 km away?
Thommes et al. (2001) ran simulations and found that, at optimal conditions (namely, a planet of ~ 10 Earth masses), migration can be complete with ~ 100,000 years. Note that this was done before in-depth research was done on the Nice model, which is very similar. However, the mechanisms are different, as are the planet masses. The difference in timescales is dramatic.
Levison et al. (2007) did explore their own model - the Nice Model. They found that it took 60 million years to 1.1 billion years for Jupiter and Saturn to break their resonance. The period of encounters and scattering lasted for 878-885 million years, followed by a period of eccentricity damping lasting for 0.3 million years to 4 million years.
What was relatively quick was the ejection of the hypothetical 5th giant planet. The change in the other giants' orbits, however, was not.
So yes, planetary migration on this scale takes a long time. A very long time.
For some really interesting results, see the graphs of semi-major axis vs. time from the various four-, five-, and six- planet models of Nesvorný & Morbidelli (2012). There are some incredible oscillation among the orbits of Uranus and Neptune in some of the simulations, which is eventually slowly damped.
Fig. 14.— Orbit histories of the giant planets in a simulation with five initial planets. See the caption of Fig. 1 for the description of orbital parameters shown here. The five planets were started in the (3:2,3:2,2:1,3:2) resonant chain, Mdisk = 20 MEarth and B(1). The fifth planet was ejected at $t =$ 6.1 Myr after the start of the simulation.
There's another relevant passage, if you're looking at setups with six giant planets:
The instability typically occurred in two steps, corresponding to the ejection of the two planets. Sometimes, as in Fig. 18, the ejection of the two planets was nearly simultaneous, but most of the times there was a significant delay between ejections. This was useful because the first planet’s ejection partially disrupted the planetesimal disk and reduced its capability to damp e55, which was then excited by the second planet’s ejection. While this mode of
instability can be important, we would need to increase the statistics (>100 simulations for each initial condition) to be able to properly resolve the small success fractions in the six-planet case.
There is no observational evidence that the sun is a member of a binary (trinary, or more) star system, where "star" means an object that is at least ~80 times the mass of jupiter and emits energy/light via standard hydrogen fusion.
Some evidence that people point to is that the majority of stars in the Galaxy (perhaps 60% or so) are binary. However, that does not mean that we are a binary system, just that we are among those which are not.
There is the Nemesis star idea which was a hypothetical binary companion to the sun to account for periodic mass extinctions roughy every 26 million years (as in, the two stars would therefore be on a 26 million year period, so the companion is a few light-years away). There are two problems with that: First, that a 2010 reanalysis of the data showed that it was too periodic over the past 500 million years, during which time we've gone around the galaxy twice, and so our hypothetical binary orbits should have been perturbed and not be perfect (though others say that the precision of the geologic timescale isn't good enough to say this). But, second, we can see stars out to a few light years -- we can see small, faint stars out to thousands of light years. We have all-sky infrared surveys that should be able to pick up the faintest even star-like objects to at least a few light-years, and yet ... nothing.
The other problem is what LDC3 pointed out in their answer: We should see some systematic motion of our own star in orbit around the common center of mass of the hypothetical binary. This would not be a yearly wobble, but rather it would be a slow motion of the entire sky on top of the other >1 year motions that we see. We now have very accurate astronomical records dating back at least a century of star positions, especially close stars. Even if we were on a 26 million year orbit - and especially one much shorter as some people claim - we should see effectively that our star is making a small arc, a part of a circle as it orbits the center of mass of it and the binary. We don't. There does not seem to be any systematic signal in the motions of stars that require the binary star model.
So, to summarize: A binary companion simply lacks any observational data; if it existed, we should not only have been able to see it by now, but we should also observe not only the stars in our sky show a systematic motion due to our orbit around it, but we should again see THAT binary star also move very quickly, relative to other stars, as it orbits the common center of mass.
To expand a bit on James Kilfiger's great answer:
The wavelength of the cosmic microwave background (as well as any other photon) is stretched proportionally to the scale factor $a(t) = 1/(1+z)$ ("the size") of the Universe, where $t$ is the age of the Universe and $z$ is the redshift of light emitted at $t$ and observed today.
Thus, at redshift $z$ the peak of the CMB spectrum was lying at a wavelength
$$
lambda_mathrm{max}(z) = frac{lambda_mathrm{max,0}}{1+z}.
$$
Today, the CMB spectrum peaks at $lambda_mathrm{max,0} simeq 0.2,mathrm{cm}$. If we take visible light to lie in the range 400-700 nm, the maximum of the CMB spectrum must have been in the visible range at redshifts
$$
z ,,=,, frac{lambda_mathrm{max,0}}{lambda_mathrm{max}(z)} - 1 ,,=,, frac{0.2,mathrm{nm}}{{4,verb+-+,7}times10^{-5},mathrm{nm}} - 1 ,,sim,, {2700,verb+-+,4700}.
$$
This can be converted to an age of the Universe by integrating (numerically) the Friedmann equation. Assuming a Planck 2015 cosmology, I get that the peak of the CMB spectrum was in the visible range when the Universe was between roughly 50,000 and 125,000 years old.
However, as James Kilfiger also mentions, even though the peak lies outside the visible range, a fraction of the spectrum is still inside. Its color depends on the response of the cones of the human eye and follows approximately the so-called Planckian locus in the CIE 1931 color space:
In the plot above, the locus traces the color of the Universe as a function of temperature (thin numbers), and age of the Universe in kiloyears (kyr; bold numbers).
That is, the Universe started out as bluish (when $Tgtrsim10^4,mathrm{K}$), became whitish at an age of around 200,000 years (when $Tsim5verb+-+6000,mathrm{K}$), and then gradually went over orange and red at $tsim1,mathrm{Myr}$ before fading into the infrared. This evolution is completely analogous to the colors of stars of a given temperature.
The ages have been calculated with Python's CosmoloPy.distances.age(), which I think is not very accurate at such small ages, but it does give the approximate age. Also, whether or not the CMB was visible to the naked eye — i.e. without a telescope to enlarge the light-collecting area — depends on the intensity at a given point.
The sun by far occupies the largest part of the solar system's mass. The mass of the sun happens to be approximately $(1.98855±0.00025)×10^{30}$ kg. Which is immensely huge. By comparison the inter-planetary medium though may occupy a large volume is nowhere dense enough to even compare to the mass of the sun which is approximately 99% of the solar system's mass.
The mass of all planets combined by comparison happens to be approximately $2.6634×10^{27}$kg.
As far as interplanetary matter is concerned, its density is to low and its volume is pretty hard to define so mass remains a highly variable quantity.
Moons and asteroids account only for a fraction of the mass of the solar system though they happen to be present in large numbers.
It all comes down to the brightness of objects (not their size). For all intents and purposes we can assume that the most distant galaxies and the small, but much closer, objects in the Oort clouds are unresolved point sources.The Oort cloud objects are too faint to see, with JWST, but it should be able to see bright galaxies and quasars even at 13 billion light years.
JWST will have a magnitude limit of something like 30th AB magnitude at red and near-infrared wavelengths. It turns out that quasars at Z~10 can have brightnesses that enable them to be detected at these magnitudes and brighter.
OK, but now work out how bright an Oort cloud object is seen in reflected light from the Sun. Taking a solar luminosity and calculating the flux at say the nearest hypothesised Oort cloud objects at about 2000 au from the Sun, we get $1.4times 10^3/(2000^2) = 3.5times10^{-4}$ W/m$^2$. Now let's be generous and say that as much as 10 per cent of this is reflected (a very high albedo) and that it is a big Oort object with a radius of 20 km. If this acts as a Lambertian reflector, then it will receive $3.5times 10^{-4} times pi times 20000^2 = 4.4times 10^{5}$ W and it will radiate $4.4times 10^{4}$ W isotropically back into a hemisphere towards the inner solar system.
Now reversing the calculation. After this light has travelled back 2000 au to the JWST, the flux received will be $4.4 times 10^4 /(2pi [2000 times 1.5times 10^{11}]^2) = 4.9 times 10^{-25}$ W/m$^2$.
We can compare this with the original flux of the Sun at the Earth. This is $1.4times 10^{3}$ W/m$^2$. Now assuming that the reflection doesn't change the solar spectrum, we can take the ratio of the solar flux at earth to the reflected light from the Oort cloud object to calculate that the Oort cloud object is 68.6 magnitudes fainter than the Sun as seen from the Earth (or close to the Earth - JWST will not be in orbit around the Earth) in any magnitude band you care to choose.
So in the near infrared, the Sun has an apparent magnitude of around -27.5. This means that the Oort cloud object would have a near infrared apparent magnitude of 41 and would thus be 11 magnitudes too faint to be seen by JWST. (And remember we chose a big object that was as close as an Oort cloud object is expected it to be and gave it a high albedo),
Now consider that 30th mag detection threshold. For a solar-type spectrum this equates to a flux threshold of $sim 10^{-20}$ W/m$^2$. Ignoring the cosmological nuances of proper distance etc., then something at 13 billion light years would have to have a luminosity of a mere $10^7$ solar luminosities to produce this kind of flux - the brightness of a modest galaxy. Of course there is also the extreme redshift to consider, so this calculation is way too optimistic, but even allowing for an order of magnitude or two, one should easily be able to see the brightest galaxies and quasars at 30th magnitude.
I wrote an experiment in HTML5 and Javascript to try to show what is happening to the curvature of spacetime as a body moves through a region.
As time progresses, I'm using the g=Gm/r2 equation to move the spacetime grid toward the object. When a grid point is within the body, I use g(1-h/R) to modify the bending according to the acceleration at the surface. This assumes uniform density.
The result is that it takes a very large mass and very long time for any sort of noticeable curvature to occur (as you'd expect). By decreasing the Earth's radius by a factor of 3, I was able to see the curvature very well as in the screenshot below.
What occurred to me then is that the the warping of space would remain long after the body of mass had moved on.
Does this warping in space ever return to truly "flat" or is the fact that we almost never orbit through the same absolute region of space mean we never experience this existing warping?
If light went through this "wake" would it bend according to the spacetime curvature created by the long distant planet? Or have I gone wrong in my model?
This demo uses real values except that the radius is shrunk by 3x to increase its gravitation effect.
earth.radius = 6.371e6 / 3;
earth.mass = 5.972e24;
spacetimeGrid.extent = 4.5e7;
You can run the demo here: https://dl.dropboxusercontent.com/u/2236585/spacetime/index.html
We are situated inside the galactic disk. The first image below, showing our approximate location, is a fabrication rather than a photograph. Such a photo would require a space probe to be tens of thousands of light years above the plane of the disk, which is patently infeasible with current technology. We don't even know exactly how many spiral arms our galaxy has!
But from where we are, we can still see a great density of gas and dust stretching 360 degrees across our sky, as you look along the galactic plane. These "inside views" are indeed real photographs - much the same as if you go outside on a dark night, well away from the city lights, you yourself will see its faint, dusty outline much like in the second image (which I can assure you is real).
I live in an 'L' shaped building on the 14th floor, so I have a pretty clear view of the horizon. Up until March 20th of this year, the sun was setting behind the other wing of the building. My living room window faces roughly NW, and on March 21st, I was able to see the sun at the horizon for the first time this year. Each day, the sun will set a little further northwest until it's setting almost directly opposite my window. Then it will begin to retreat until one day in September, when it will begin to set behind the other wing of the building again.
This is the first year I took note of the day that sunset was first visible, and I'm planning to note which day is the last day I can see the sun set.
According to wiki page, the mean temperature of Pluto is 44 degree K (-229 degree C). However, given that Pluto is tilted greatly and there is daylight for long periods on part of Pluto facing the Sun and also given that atmosphere of Pluto scatters light to a high degree, the surface temperature on that part of Pluto may be higher. The atmosphere of Pluto contains methane which is a powerful greenhouse gas, and that also could push up the temperature.
What can be the maximum temperature on part of Pluto facing the Sun? Does available temperature estimate take into account findings of New Horizons probe which flyby Pluto earlier this year? Thanks for your insight.
"White" stars are typically much brighter than Red stars, as both the "color & brightness" of a star are directly proportional to the temperature. The only reason there are "bright" red stars is that their radius is incredibly large. Note that the "color" of a star is directly linked to the temperature.
The equation that best demonstrates this is the luminosity equation of a black body. Stars aren't perfect black bodies, but they are close enough that they are treated as such.
L = 4πR²σT⁴
this equation tells us that the Luminosity (L) is proportionate to the Radius Squared (R²) and the Temperature to the Fourth power (T⁴). The bigger the brighter, or, the hotter the brighter. Meaning that for a given radius the hotter the star, the more luminous, and the same goes for stars of the same temperature, the larger the radius the more luminous.
White dwarfs on the other hand are not stars in the sense that they do not fuse anything, they simply glow due to the lingering heat that was generated during their time as stars.
As shown in the HR-Diagram, White Dwarfs are some of the hottest objects in the universe, and as stated by agtoever there has not been enough time for even the oldest white dwarf to have cooled passed something like 4800K.
Large telescopes don't use a single CCD; they use arrays of them. Since CCDs can't be seamlessly joined, this means that every captured image must have gaps. But, the resulting images almost never have gaps when they are presented to the public.
How are these gaps filled? Does the telescope capture a number of overlapping images and then merge the results? Wouldn't this result in differing exposure levels for differing areas of the image, which, even though it could be compensated for in the processing, would give different noise characteristics to the different areas?
But if yellow light is what's getting through the atmosphere - why doesn't everything have a yellow tinge to it?
Everything does have a yellowish tinge to it, but only for roughly the hour after sunrise and the hour before sunset. These a photographer's golden hours:
As the Sun rises, its color changes from reddish to orangish to yellowish -- and then to nearly white, even when the Sun never does get very high.
You must not (and cannot) look at the Sun, even for the most fleeting of moments, once the Sun gets above 10 to 15 or so degrees above the horizon. The only time you can look at the Sun, even briefly, is when the Sun is low on the horizon and air mass is high. We think the Sun is yellow (or in some civilizations, orange or even red) because we cannot see the Sun for what it truly is, which is nearly white.
What bit of off-white color sunlight does have when the Sun is 10 to 15 or so degrees above the horizon is offset by the blue light from the sky, aka "diffuse sky radiation". The color of the light from the Sun and the sky when the Sun is high in the sky is about as white as light can get.
There's a bit of physiology going on here as well. To those who have lived their most of their lives inside buildings, bright sunlight is a bit yellowish and a bit too harsh. To those who spend much of their time outdoors, the fluorescent lighting that now dominates the urban landscape is a bit bluish and a bit too harsh. What city dwellers who rarely ventures outdoors thinks of as white light is not very white to those who spends most of their waking hours outdoors, and vice versa.
I would have added this as a comment (not enough rep yet, I'm afraid)...
To elaborate on Andy's answer, the first reason is that the surface of the satellite dishes are too coarse to form any kind of image. Polished optical surfaces are smoothed to a polish (generally much smoother than one would achieve in polishing a car, though).
Other problems come from the precision of the form - the paraboloid of the satellite dish may have errors on the order of millimeters, because radio waves (which satellites use) don't care about that much error. In fact, satellite dishes are EXACTLY the same as radio telescopes, except the receiver cone is designed for frequencies used by communication satellites instead of frequencies of interest to astronomy.
Just for your interest, old satellite dishes ARE well-suited to function as rough solar collectors (for a solar oven) or as an acoustic telescope (position a microphone or speaker where the receiver cone normally lies)!
Here are some thoughts adapted to an answer I placed on Phyiscs SE to a similar question some time ago. In order to observe the past we need to detect light from the Earth, reflected back to us from somewhere distant in space.
The average albedo of the Earth is about 0.3 (i.e. it reflects 30 percent of the light incident upon it). The amount of incident radiation from the Sun at any moment is the solar constant ($F sim 1.3 times 10^3$ Wm$^{-2}$) integrated over a hemisphere. Thus the total reflected light from the Earth is about $L=5times 10^{16}$ W.
If this light from the Earth has the same spectrum as sunlight and it gets reflected from something which is positioned optimally - i.e. it sees the full illuminated hemisphere. then, roughly speaking, the incident flux on a reflecting body will be $L/2pi d^2$ (because it is scattered roughly into a hemisphere of the sky).
Now we have to explore some divergent scenarios.
Let's be generous and say it is a perfect reflector, but we can't assume specular reflection. Instead let's assume the reflected light is also scattered isotropically into a $2pi$ solid angle. Thus the radiation we get back will be
$$ f = frac{L}{2pi d^2} frac{pi r^2}{2pi d^2} = frac{L r^2}{4pi d^4},$$
where $r$ is the radius of the thing doing the reflecting.
To turn a flux into an astronomical magnitude we note that the Sun has a visual magnitude of $-26.74$. The apparent magnitude of the reflected light will be given by
$$ m = 2.5log_{10} left(frac{F}{f}right) -26.74 = 2.5 log_{10} left(frac{4F pi d^4}{L r^2}right) -26.74 $$
So let's put in some numbers. Assume $r=R_{odot}$ (i.e. a reflector as big as the Sun) and let $d$ be 1000 light years. From this I calculate $m=85$.
To put this in context, the Hubble space telescope ultra deep field has a magnitude limit of around $m=30$ (http://arxiv.org/abs/1305.1931 ) and each 5 magnitudes on top of that corresponds to a factor of 100 decrease in brightness. So $m=85$ is about 22 orders of magnitude fainter than detectable by HST. What's worse, the reflector also scatters all the light from the rest of the universe, so picking out the signal from the earth will be utterly futile.
How did it get there? Let's leave that aside. In this case we would just be looking at an image of the Earth as if it were 2000 light years away (assuming everything gets reflected). The flux received back at Earth in this case:
$$ f = frac{L}{2pi [2d]^2} $$
with $d=1000$ light years, which will result in an apparent magnitude at the earth of
$m=37$.
OK, this is more promising, but still 7 magnitudes below detection with the HST and perhaps 5 magnitudes fainter than might be detected with the James Webb Space Telescope if and when it does an ultra-deep field. It is unclear whether the sky will be actually full of optical sources at this level of faintness and so even higher spatial resolution than HST/JWST might be required to pick it out even if we had the sensitivity.
Of course this doesn't help you see into the past because we would have to send the telescope there. But it could help those in the future see into their past.
Assuming this is technically feasible, the Earth will have a maximum brightness corresponding to $m sim 35$ so something a lot better than JWST would be required and that ignores the problem of the brightness contrast with the Sun, which would be separated by only 0.03 arcseconds from the Earth at that distance.
Note also that these calculations are merely to detect the light from the whole Earth. To extract anything meaningful would mean collecting a spectrum at the very least! And all this is for only 2000 years into the past.
Amphibians are produced externally so they do not get proteins and ATP from the mother during their maturation and differentiation. This means that they must have some way to get those things to protect themselves and survive.
The way that they get these things is yolk. Amphibians start to have yolk-filled endoderm during blastulation, [page 9, Gilbert, Developmental Biology].
I am considering the frog here as an example of amphibian. To know that things need energy to survive and that they must get it somewhere, you can deduce this solution. So nutrition and external "hatching" or development outside of female is the key here.
The overall answer to the question:
The organism must have a yolk sac, since it has yolk.
The second question seems to be an idealized one.
Without blastocoele but with yolk-sac suggests me that the blastocoel has developed already to yolk-sac. If that is the case, then I would call the thing gastrula, since for instance some species' ectoderm develops into yolk-sac. I assume that the question refers with "yolk-sac" to complete yolk-sac, not to the developing one during gastrulation.
If it is about developing yolk-sac, I would say that the cell is still blastula but is developing to gastrula. The yolk sac is not ready until gastrulation is complete.
Tonight (10:00 PM EST Middlesex County, Mass.) when looking at the mostly clear sky I observed what I at first believed was a meteor. This point of light was moving east to northeast and that would be in the vicinity of the Lyrid meteor shower. In the app Star Walk for iOS it says that the "n-Lyrids" meteor shower peaks on May 9th. However information I found online about this meteor shower says it already peaked back in April. (I did see a bright meteor a couple weeks ago.) Can anyone offer insight on possible meteors right now? I highly doubt that this was a meteor though, because it lasted way too long, probably two minutes. My next guess was the International Space Station but NASA's Spot The Station did not show any viewing times for my area today. The Star Walk app also showed the ISS well below the horizon. This leads me to conclude the object was a plane, however there were no blinking lights and the only time its light changed was when it passed behind a thin cloud. Just did some more research and it turns out it could have been satellite glint such an Iridium flare. I did not see the beginning of the light's appearance, so it is possible I missed the flare. However flares do not last for minutes, but glint from solar panels or other satellites may last this long. This is likely what I say. If anyone has had similar experiences, any information would be appreciated.
Has there been any study to suggest that either quasars of gamma ray bursts are more detectable from further distances in the universe?
I've read that quasars are the most powerful energy in the universe but gamma ray bursts are detectable from further distances. Logically one would think that if some entity's energy were detectable from further distances than another entity's energy, then it'd be the entity with more energy than the other.
I'm just looking to see if anyone has any definitive study references that suggest why either quasar or gamma ray bursts have more energy than the other and to explain why if true the one with less energy is visible from further distances than the one with more energy.
I'm not really an astrophysicist type guy but I do some reading and watch different scientific shows that interest my since a lot is still unknown, so I'm hopeful someone here more familiar with this can answer or clarify this for me or give me some references to anything that gives clue to any of this.
I apologize for not having direct reference to any readings I've read that bring about this question in my mind but I know I read something that said gamma ray burst are detectable from further distances than any other energy entity and I've also read quasars are the most powerful energy sources in the universe, so I'm just looking for some clarify on this topic whether right or wrong with anything further you can provide.
