# Tag Info

22

First, the speed of other galaxies isn't too helpful. For example, the radial velocity of the Andromeda galaxy relatively to us is 300 km/s, i.e. 0.1% of the speed of light only. Moreover, internally, everything in that galaxy moves by pretty much the same speed and is confined to the vicinity of that galaxy which makes us pretty sure that no piece will ...

21

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 ...

20

Why don't we observe any relativistic asteroids? The answer to this question would not be complete without mentioning the virial theorem. Considering our galaxy as a system of $N$ gravitating objects, according to the virial theorem, twice the average total kinetic energy of all objects, plus the average total potential energy of these objects, adds up ...

16

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 ...

11

Did you read the Wikipedia article? It explains the signal rather well, I think. At any rate, it is called the Wow! signal because, as the picture shows someone wrote Wow! in the margin. As for the code and why they were excited, I quote the Wikipedia article, The circled alphanumeric code 6EQUJ5 describes the intensity variation of the signal. A ...

10

The angular resolution of a telescope is approximated by the formula: $$\sin \theta \approx\theta\approx1.220\frac{\lambda}{D}$$ So, if we know the angle, we can calculate the diameter $D$. The rest really depends on how big the flag is and where on moon it has been planted. Assuming 50$\text {cm}$ as the diameter of the flag, and assuming the flag 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 ...

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 ...

9

You mean like Arthur C. Clarke's 2010 when Jupiter turns into a star? We often turn to Jupiter's mass ($M_j$) when thinking about this problem. It turns out there's a whole class of stars that fuse so faintly that we can only see them well in infrared. Brown dwarfs (which are still called "stars") turned out to be so cool that only new infrared ...

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 ...

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

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 ...

7

If the majority of the radiation emitted by a star is infrared, the majority of the visible light emitted will be red. We don't see the lower half of the spectrum the star emmits If the majority of the radiation emitted by a star is infrared but some is visible, the average visible light emitted will be yellow. Little blue is emmitted and any green ...

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

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

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

Century First, you'd have to watch through a night to see if Polaris wobbles - currently, the radius is about 1° I think, but that changes with precession (and nutation, but that's small enough to ignore). Once you know that, you can try to find a point in the sky that stays still all the time (like Polaris nearly does in our time). This is celestial pole, ...

6

Red giants and asymptotic giants have some close similarities, and one actually evolves into the other. Both have an extended envelope of relatively cool, non-burning material (mostly $\rm{H}$, $\rm{He}$). They also each have a core of dense, non-burning material; in the case of the red giant this is mostly $\rm{He}$, while for the asymptotic giant it's ...

6

While the chosen answer isn't incorrect it doesn't really answer the question -- that it isn't that rare on other planets. For a total eclipse you have to fall into the Umbra portion of the shadow. From this image you can see that the size of the moon and the distance from the sun as well all play an important role. If we look at the planets and moons ...

6

A galaxy spectrum is a quite complex and complicated topic, and many entire careers are fully devoted to understanding them, so this can only be a simplified answer. It is still quite lengthy, though, so if you're impatient, I've summarised it at the bottom. A blend of starlight of different spectral types makes up the continuum. The light is emitted by ...

6

There's actually at least one very big clue that's been accessible to skygazers since the earliest times: the first quarter moon at dusk. Every child in the northern hemisphere going back to 30,000 BCE likely would have been familiar with how 1st-quarter moons always tend to rise at noon, reach its highest point at sunset (with an azimuth directly south), ...

6

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 ...

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

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 ...

5

This is likely stretching your requirements a little bit, but I find this related method ingenious and surprising. You can actually find your longitude as well as the other data from @Pharoh's answer if you can remember very well how the night sky looked at your original position on Earth. The method is the Lunar Distance Method (see Wiki page with this ...

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

You can use the center of mass formula. Set the origin of your coordinate system at the center of the Earth, then $\vec{r}_1 = \vec{0}$ and $\vec{r}_2 = d$ and $$r_{center} = \frac{m_1r_1+m_2r_2}{m_1+m_2} = \frac{m_2}{m_1+m_2} \cdot d$$ as you have as well.

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