# Tag Info

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 ...

22

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 ...

17

This is not possible. The lowest possible mass for a main sequence star (sustaining H-1 fusion; it's the regular kind of star) is around 80 Jupiter masses. Just below this, objects are referred to as Brown Dwarfs, which are technically not stars. Whereas the highest possible mass for a terrestrial planet is about 5-10 Earth masses (as per here). Above this ...

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

Is it possible for a star to have the same mass and radius as e.g. the Moon and orbit a planet like Earth at the same distance (at which Moon orbits Earth in actuality)? No. The lowest mass type of star is a Brown Dwarf, which still has a mass greater than that of Jupiter. Even brown dwarfs have too little mass to fuse light hydrogen. Neutron stars ...

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 ...

9

It is happening because of the acceleration of the Earth orbital speed around the Sun (Earth is near the perihelion). Between December 13 and December 31 the Earth is speeding up and also it is normally rotating around its axis. These 2 movements (constant rotation and increasing orbital speed) add up to create the observed apparent movement of the Sun on ...

9

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 ...

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

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

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 ...

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

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 ...

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

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

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 ...

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

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 ...

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 ...

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

The Earth receives approximately $6.8\text{mW/m}^2$ of reflected sunlight from the moon (see below for details of how I calculated that). However, the sunlight is also absorbed by the moon and this raise the surface temperature. So the moon also emits thermal radiation towards the Earth (assuming the highest day time temperature of 400K, see comments below ...

5

There is indeed an optimal angle where Venus is brightest. Have a look at the following figure: The distance between Earth and Sun is $\Delta$, between Sun and Venus is $r$, and between Earth and Venus is $\rho$. The amount of light that Venus receives from the Sun is $$f \sim \frac{\pi a^2}{r^2},$$ where $a$ is the radius of Venus. When viewed from ...

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 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 ... 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 Orbits are a funny thing, in space bodies have gravitational effects on one another. In a system like our solar system none of the objects including the sun are truly stationary in relation to one another. Even stars move around based on the gravitational pull of their planetary bodies. this wobble allows us to detect planets in solar systems far away from ... 5 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? 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

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 ...

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