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Although the photon appears to exist without physical volume or geometrical size, we can measure the region where the wave's magnitude is non-negligible. This happens at about half a fermi, or roughly $0.5*10^{-15}$ m.


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You have to define what you mean by the "surface". Conventionally the optical photosphere of a star is defined in term of its optical depth to radiation - usually the photosphere is said to be where the optical depth reaches 2/3. Clearly, the gas at this optical depth has a pressure - and this is what one would normally talk about in terms of the surface ...


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The gasses in a stellar surface (photosphere) are radiating, and their spectrum includes both thermal (doppler shift, black-body brightness dependence) and pressure (linewidth broadening) effects. It is possible to determine pressure and temperature from the spectrum, at the photosphere and perhaps (with some difficulty) above the photosphere in the ...


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Stars don't have a well defined surface. If you plot the density as a function of radial distance then it falls smoothly with distance and in principle is non-zero out to very large distances. A star is basically a ball of (ionised) gas. If you take another example of a ball of gas (well, a spherical shell of gas), i.e. the Earth's atmosphere, it doesn't ...


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Yes of course. There are techniques which allow astronomers and astrophysicists to calculate these quantities quite efficiently and correctly and it depends on the spectrum, the luminosity and the distance of the star. In fact, from just the luminosity and the distance, one can compute almost everything important about the star including its age, ...


2

Why are Lunar Eclipse more common than Solar Eclipse? They aren't. Lunar eclipses and solar eclipses occur with almost equal frequency. From http://eclipse.gsfc.nasa.gov/eclipse.html and pages within, there were / will be 11898 solar eclipses of all types and 12064 lunar eclipses of all types in the five millennia between 2000 BCE to 3000 CE. Lunar ...


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The apparent size of the moon (from Earth) is about the same as the sun's. Therefore the moon has to pass over the sun's path at the time of the new moon for there to be a solar eclipse. The apparent size of Earth (from the moon) is much greater than that of the sun - because the Earth's diameter is about 4 times that of the moon, and hence the alignment ...


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Absolutely it can - and it happens all the time. If you excite an atom, it can go through various "stages" of decay back to the ground state - with each drop in energy resulting in an emission of radiation. This happens during photosynthesis: see this page from which I copy this image: As you can see, there are multiple paths for the energy to be lost by ...


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The answer is "Yes" but not the way you might expect. It is possible to construct a telescope mirror from rotating liquid metal.Mercury used to be used but something like Gallium is safer and better. So print a cradle for it, put in the Gallium, raise the equipment past the melting point (about 30 degC), spin gently to get a parabolic surface, and then ...


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Here's how to measure stellar aberration: Take a telescope and point it straight up. Attach it to something massive and steady, like a chimney stack. Let the scope act like a pendulum. It's fixed at the top, and moves at the bottom. The plane in which it swings is the local meridian. That is, it only swings north-south, not east west. With this setup, you ...


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How long time does it take before three planets achieve the same relative position? The answer is never, except for the case when their orbital periods can be expressed with low integers, like the 4:2:1 resonance of Io, Europa and Ganymede However, what you are asking about is when they are going to be in almost the same position again, a quazi-period. To ...


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SN 1987A is (was?) in the Large Magellanic Cloud, which is gravitationally bound to the Milky Way. This means that its motion relative to us is only minimally affected by cosmological expansion, and talking about it in terms of a $z$ parameter is misleading at best. The best estimates of the distance to SN 1987A are about 168,000 light-years. If you ...


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Of course you won't find it anywhere - SN 1987A is in the Large Magellanic Cloud, just 168000 ly away. At this scales cosmological expansion is negligible compared to other processes so measuring its redshift is says little useful about its distance. $z=0.1$ corresponds roughly to 1 Gly. The universe is HUGE. Here's plot from wikipedia


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The Magellanic clouds are satellite galaxies of the Milky Way. They are right next door. Google says 61 kPc to the LMC which means trivial cosmological redshift.


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Without knowing much about Jean's instability I can probably help you with the derivation of the expression. You start with the expression in 1) and expand it to first order in $\delta P$ and $\delta R$. Then you use that $R_0$ and $P_0$ are stationary solutions i.e $\ddot{R_0}=0$ thus $GmM/R_0^2= 4 \pi R_0^2 P_0$. You can use this to simplify the ...


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So this is really really easy and I don't know why the top answer doesn't explain what's going on. I have a lamp on my desk, which is opposite the headboard of my bed. Sometimes I lie there reading with the lamp on to illuminate the room, but with my lowered vantage point, oh no, a first world problem hits - there's a bright light in my field of vision ...


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It's a combination of a few things. Firstly when we are looking for exo-planets we know we are not going to observe them based on their luminoscity, therefore we use different techniques based on how the exo planet will effect the light we observe from their sun. This method works brilliantly if the star and exoplanet are in relatively close proximity but ...


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Pretty simple reason really. We only see exoplanets under extremely lucky circumstances. So we are only seeing a tiny tiny fraction of all exoplanets. If for example we are only seeing 0.1% of all exoplanets in each star system we look at, that is a HECK of a lot worse than the 8 out of 9 in our own star system.


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The reason why we can see exoplanets 13,000 light years away but not a planet 200 AU away (about 30 light-hours) is because these planets are found using different techniques. The planet discussed in the article I linked was discovered using a technique known as "microlensing," which requires a star to pass behind another star with a planet around it. The ...


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The problem with finding a new planet in our solar system is not that it is too faint, but knowing where to look in a big, big sky. This putative planet 9 is likely to be in the range 20-28th magnitude. This is faint (especially at the faint end), but certainly not out of reach of today's big telescopes. I understand that various parts of the sky are ...


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We haven't detected planets millions of light years away. Right now the most distant is less than 20,000 light years away. Even for the planets we have detected, they are for the most part not "seen" or imaged directly. Instead they are found by the effect they have on the parent star (usually gravitational wobble or transit detection). In both cases, ...


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There is an intriguing possibility that the answer is yes, in a very fundamental way! All amino acids (except glycine) come in two chiralities, left-handed and right-handed, related by parity. However, all amino acids used in living beings are left-handed. Evidently, by chance, early in Earth's history, left-handed compounds gained an advantage somehow. ...


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You would most definitely use Kepler's third law. With almost all planets, the mass of the planet is negligible so it is ignored. We know the orbital period and the mass. Use those and Kepler's third law to write an equation for the orbital radius.



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