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67

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 ...


23

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 ...


8

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 ...


6

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, ...


5

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.


3

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 ...


3

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.


3

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 ...


3

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, ...


3

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 ...


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 ...


2

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 ...


1

The practical limit in terms of sky coverage is probably given by funding. The HDF, according to wikipedia, covers a 24-millionth of the entire sky. We would have to launch thousands of Hubble-like instruments to cover the entire sky at that resolution and sensitivity. Much of that area is covered by the Milky Way and and gas clouds, anyway, so not all deep ...


1

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. ...


1

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 ...


1

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 ...


1

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


1

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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