# Tag Info

## Hot answers tagged astronomy

29

The highest resolution 3d printers I know of are around 1600dpi, which is a resolution of about 15$\mu m$. Telescope mirrors have to be smooth to fractions of a wavelength of light, so the resolution of current printers is nowhere near good enough. Whether 3D printers could one day be good enough is a different question, but given that the improvement in ...

23

A typical giant galaxy, such as the one you've provided a picture of, has a radius of something like $10\;\rm kpc$ (kiloparsec - $1\;\rm pc \approx 3.2\;ly$). A supermassive black hole hosted in such a galaxy has a mass of something like $10^6-10^9\;\rm M_\odot$ (solar mass, $1\;\rm M_\odot \approx 2\times10^{30}\; kg$). The monstrous billion solar mass ...

23

One cannot tell by the light spectra. Hydrogen and antihydrogen would give the same lines in the spectrum. The prevalence of matter over antimatter from other evidence indicates matter is predominant in the observable universe, and here is a nice review. How do we really know that the universe is not matter-antimatter symmetric? The Moon: Neil ...

11

Generally speaking they refer to the distances from us when the light as emitted. No correction is usually made to say how far away the object is from us now, because this correction would be very small and inconsequential compared to the uncertainty in the original distance measurement. For instance, taking the Andromeda M31 galaxy as an example. Riess et ...

11

Measuring $w$ is actually what I do for a living. The current best measurements put $w$ at $-1$ but with an uncertainty of $5\%$, so there's a little room for $w \ne -1$ models, but it's not big and getting smaller all the time. Indeed, we'd all be thrilled if, as measurements got more precise, $w \ne -1$ turns out to be the case because the $\Lambda$CDM ...

10

Surprisingly it's quite easy to answer this because we can use the cosmic microwave background as a reference. The CMB gives us an average inertial frame for the universe so our motion relative to it is the closest we can come to defining the Solar System's motion through space. The CMB is isotropic, but because we are moving relative to it the radiation is ...

10

Increasing the diameter and distance of the Moon by a factor 2 would lead to a number of very subtle differences. I will list the ones I came up with: Apparent size If $R_m$ is the radius of the Moon and $D_m$ its geocentric distance (that is, the distance between the centre of the Moon and the centre of the Earth), then its geocentric angular diameter is ...

8

The short answer is you can't, or at least not at all easily. Your detector has only a single detection plane, and almost all muons are minimum ionizing, so you get essentially the same energy deposition from every muon (well, there is a factor from the angle of incidence the detection plane). The usual mechanism for measuring the energy of a particle are ...

8

Stars Indeed, the most readily apparent observables for stars are (1) their apparent luminosities, and (2) their spectra (or even just colors if you can only do photometry). The age has to be inferred, and this is where modelling comes into play. The Vogt-Russell "theorem" is the assumption that the initial mass and chemical composition of a star uniquely ...

8

According to E. V. Petjeva (2011), the measured rate of change of the Earth-Sun distance (astronomical unit) is (1.2 +/- 3.2) cm/yr, with the uncertainty value representing 3 standard deviations. In other words, any change is within the uncertainty of the measurement. She specifically addresses the Krasinsky and Brumberg value. E. M. Standish has also ...

7

While most measurements in astronomy are better in space, precision spectroscopy can actually do quite well on the ground. One of the best spectrographs (some would say the best) is HARPS, the High-Accuracy Radial Velocity Planetary Searcher used for finding extrasolar planets. As described in its instrument paper (pdf; note that the sole purpose of this ...

7

If the universe is flat does it mean it doesn't exist? What kind of incoherent question is this? "If the universe is flat" presumes the universe exists and is spatially flat. Your question amounts to "Does the existence of a spatially flat universe mean the non-existence of a spatially flat universe?". Isn't the answer analytically no?

7

Compared to naked eye view, a telescope image never increases surface brightness. This fact is related to the concept 'etendue'. However, although the image formed on your retina is never brighter than the corresponding naked eye image, the image through a telescope is magnified. This means that looking through a telescope at the sun can expose your whole ...

7

The question is dealt with in some detail in this article by John Baez. Although the article assumes only a basic understanding of physics it's probably a bit too much for the non-physicist so I'll summarise. As a gas cloud collapses the particles within it are confined to a smaller volume of space so the entropy associated with their position (call this ...

7

To calibrate our expectations, consider the largest nuclear weapon ever detonated, the Tsar Bomba. It's yield was at most about $58$ megatons TNT equivalent, or about $2.43\times10^{24}$ erg. Now, let's consider a smallish star, something like Gliese 581, which is reasonably nearby, small and faint, and has a planetary system (of some sort: the number of ...

6

This would ultimately be more a problem of signal processing than physics. The situation is detecting a signal at a very low signal to noise ratio. At the broadband level, the noise (starlight) is several orders of magnitude more intense than the signal (the explosion). The only hope would be some sort of spectral technique , taking advantage of spectral ...

6

In this statement $\sigma = 10.2$, the uncertainty in the equivalent width. Overall, the statement means that you have a confidence of 2.2$\sigma$ that the absorption line is present (i.e. the equivalent width is 2.2 times its uncertainty bigger than zero, corresponding to a <5% chance that the absorption dip seen is due to measurement uncertainty). This ...

6

That's not quite correct. You may have noticed that during summer the days are longer (and the nights shorter) than during winter. That is because the earth's axis is tilted about $23^o$ from the plane of it's orbit around the sun. With this tilt, as the earth travels around the sun the northern hemisphere gets longer days is the north pole is tilted ...

6

You are neglecting two important facts. The first one is that stars, toward the end of their lives, return to the interstellar medium (ISM) a lot of their initial mass, but now enriched with heavy elements produced by nuclear reactions inside the stars themselves. In this way, younger stars which form from the ISM begin their life with a larger fraction ...

6

The BICEP2 paper reports a tensor/scalar ratio r=0.20+0.07−0.05, but then says: This is the value taken before corrections. There exist contributions to the B- mode due to changes in the photon polarization of the CMB while it is traveling before reaching the detector. The dust is the interstellar dust that has to be modeled. Subtracting the various ...

6

It seems from the Nature paper that what we see is a temperature variation map with the darkest and the brightest regions representing a temperature difference of $\sim 10 \%$ of the mean temperature (remember, though, that by Wien's displacement law, temperature and brightness are connected, so in some sense it is also a brightness map). The ...

5

If the diameter would be increased by a factor of 2, then if the density stays the same, the mass would increase by a factor of $2^3 = 8$, since volume is proportional to the cube of the diameter. The main effects of doing this are: the Moon's pull on the Earth would be double of what it is now, since the gravitational force is proportional to the ...

5

The Plummer model has a potential of the form $$\Phi(r)=-\frac{1}{\sqrt{r^2+1}}$$ (obviously ignoring all constants). Equating the above with the kinetic energy, you get $$\frac12v_e^2+\Phi(r)=0\to v_e(r)=\sqrt{2}\left(r+1\right)^{-1/4}$$ This velocity is the maximum velocity you can have at a radius $r$, so we must have that $0\leq v\leq v_{e}$. In ...

5

From Kepler's third law you can find that $$\frac{GM_\odot}{4\pi^2} = 1 \frac{\text{AU}^3}{\text{year}^2}$$ where $M_\odot$ is the mass of the sun. For a solar system simulation these units will be more convenient than Earth masses.

5

A brief overview of stellar evolution can be depicted in the following image: (From here which says it is originally from an encyclopedia; click here for larger image). The heavier stars (top track) have very short life times (a few million years) because they run through hydrogen, helium, carbon+oxygen, ..., iron fusion in the core. Once a particular ...

5

SSC: synchrotron self-Compton BBC: Compton upscattered blackbody radiation; Compton upscattering of stellar blackbody photons XC: upscattering of photons emitted by the accretion flow; accretion flow photons From: http://arxiv.org/abs/1307.1309 and http://arxiv.org/abs/1403.4768

5

Have a look at this article. It gives the number as $10^{24}$ rather than $10^{23}$, but it's such a vague estimate that a factor of ten is within the expected error. The number is the number of stars in the observable universe i.e. within 13.7 billion light years of Earth at the time the light we see today was emitted. Note that visible means visible to a ...

5

The inconsistency comes from assuming the planet has a greater-than-infinitesimal size while leaving the star as a point source. Usually in transit diagrams we think of the star as a disk of radius $R_\mathrm{star}$ emitting parallel rays perpendicular to its surface. The $\pi R_\mathrm{star}^2$ area of the star is partially blocked by the \$\pi ...

5

I've added this because I don't think the accepted answer is very clear. Estimating the number of stars in the Galaxy relies mostly on two things. We estimate the present day mass function (that is the number of stars that exist per unit mass per unit volume) in the solar neighbourhood. We construct a model for the overall density distribution of the ...

4

In this link, the contradiction is "explained": The tremendous expansion greatly dilutes any initial curvature. Think, for example, of standing on a basketball. It would be obvious that you are standing on a (2-dimensional) curved surface. Now imagine expanding the basketball to the size of the Earth. As you stand on it now, it will appear to be flat ...

Only top voted, non community-wiki answers of a minimum length are eligible