# Tag Info

42

The sun is the nearest star to Alpha Centauri (unless you count Proxima Centauri, which is really part of the same system). There is a very small and dim pair of brown dwarfs, called Luhman 16 that are closer, at about 3.6 light years from Alpha Centauri. Brown dwarfs are not true stars, but they do glow from their own heat. They were only discovered in ...

37

It is a matter of exposure and dynamic range. A sensor like a camera can only handle inputs in a certain range of intensities, and much of photographic skill (or smart presets) is about mapping the outside light onto this range so the details you care about show up rather than turn into white or black. If you take a picture of a brightly lit scene, in order ...

20

John's answer is correct. For a few more details: Stars brightness works out to roughly the 4th power of their relative mass. This falls off a bit for very large stars, but for smaller stars it's in the ballpark. (Red-giant phases not included.) Proxima Centauri is about 12% the mass of our sun, and about 14% the sun's diameter. At 12% its mass, its ...

15

Brown dwarfs have historically been difficult to detect (directly) simply because of how faint they are. Typical luminosities may range from $10^{-3}L_{\odot}$ to $10^{-5}L_{\odot}$ depending on spectral type. Any object that dim will be tough to find, regardless of spectral type or what sort of telescope you're using - you can have the largest, highest-...

14

Anders's answer is entirely fine, but I'd like to add some extra information. As evidenced by the transcripts, reflected Earth light is quite strong even at this distance: The earthshine coming through the window is so bright you can read a book by it. That is, even with the lights turned off, it would probably be tricky to see the stars unless you ...

12

There is no true upper bound. You could argue that a galaxy is a vast multiple star system, with hundreds of billions of stars. Natural galaxies are not perfectly stable (random encounters eject stars over very long timescales), but one can construct stable systems by taking pairs of similar-mass binaries and placing them into a remote circular orbit around ...

8

The way it works is as follows. We do detailed studies of stars in the solar neighbourhood. This establishes the local density of stars and the mix of masses they possess (called the stellar mass function). We compare that with the mass function of clusters of stars and note that to first order it appears invariant. We can then triangulate the problem in ...

7

As you've correctly mentioned observation biases play a huge role in current understanding of planetary populations. Let's have a look at a slightly outdated plot (age 2 months) that shows us the semimajor-axis vs mass distribution of currently known planets including KOI's (unconfirmed planets). Here we see two things: Hot Jupiters seem to make up a ...

7

Some classical Cepheids pulsate simultaneously in two or even three modes. Their lightcurves can be explained as a overposition of fundamental plus overtone modes. The terminology you used is also found in this paper (where you have also other references): "the discovery of many double-mode Cepheids (DMCs) pulsating in both the fundamental and first-...

7

Names are common enough for solar system bodies, but generally stars don't have names, they have identifiers from various catalogues. In the case of Pulsars. they are named by their location in the sky. Just like Earth has longitude and latitude, every point in the sky has a "right ascension" and "declination". Due to the slow movement ...

6

Exactly the difference between binary and double stars. Trinary System Three stars gravitationally bound to each other. Here are some examples Triple Star System Three stars that appear to be together but any one of them is gravitationally unbound. Here are some examples.

6

The star formation process from giant molecular cloud to unobscured protostar is thought to take about a million years. So the answer is no. Similarly, there are very few large scale physical processes that occur in the universe on a human timescale. Nevertheless we are sophisticated enough to understand that you do not necessarily have to see something ...

6

It is just a poor choice of words. The comment refers to a discovery paper by Ferguson et al. (1998), where they used a very deep image in the Virgo cluster to establish that there were an excess of distant stars compared with a control field outside of the cluster. This excess of about 630 stars were all in one tiny HST field of view, so form a "group" in ...

6

It just means that in a star that is an unresolved binary system (i.e. the light you receive comes from both stars), the secondary contributes a fraction of that light. This fraction could be expressed as a fraction of the total luminosity or a fraction of the flux at a particular wavelength or in a particular waveband. How do you work it out theoretically? ...

6

The rules from the the IAU for official designations are posted neatly here. More casual names, like the Black Widow Nebula, are not standardized, as far as I know. For astronomical objects outside of our own Solar System: the designation of astronomical objects beyond the Solar System should consist of at least two parts — a leading acronym and a sequence ...

5

There's a whole Wikipedia page about it. https://en.wikipedia.org/wiki/Solar_analog If you don't want Alpha Cen A, then 18 Sco might be the one.

5

Great question. The goldilocks zone is usually defined in terms of a region where the equilibrium temperature of the planet lies between some temperature limits (these temperature limits are somewhat debatable, but irrelevant for the purposes of this question - the boundary becomes fuzzy). This region can be calculated by working out how much flux is ...

5

Cepheid pulsations The basic description of the mechanism behind Cepheid pulsations is given here: The accepted explanation for the pulsation of Cepheids is called the Eddington valve,[38] or κ-mechanism, where the Greek letter κ (kappa) denotes gas opacity. Helium is the gas thought to be most active in the process. Doubly ionized helium (helium whose ...

5

The reason is that gravity is a radial force, and so conserves angular momentum $\mathbf{L} = \mathbf{r}\times m\mathbf{v}$, where $\mathbf{r}$ is the vector to the planet from the star and $\mathbf{v}$ is its velocity. First, remember that the cross product of two vectors $\mathbf{a}$ and $\mathbf{b}$ has a magnitude that's equal to the area of a ...

5

It's a matter of statistics. Scientists take a small amount of the space (let's say 1 second of arc). They look at it carefully with strong telescopes, and count all the stars and galaxies they see. Then, they extrapolate that number at the total visible space. Of course they can compute several spots of the space and make an average count. Since the ...

5

The tidal radius for a star or collection of stars near the Sun and on a circular orbit (matches Sirius reasonably well) is given by (e.g. Pinfield et al. 1998) $$R_T \simeq \left( \frac{GM}{2(A-B)^2} \right)^{1/3} ,$$ where $A$ and $B$ are the Oort constants and $M$ is the total mass. Using the value of $A-B$ from Feast et al. (1997), this reduces to R_T ...

5

I think you're overestimating how quickly stars appear to move as viewed from Earth. Once a telescope like Kepler identifies a solar system around a star, we know where that star is. Stars have coordinates associated with them that astronomers can use to know where to point their telescopes to see it. For example, look at the star Betelgeuse. Look in the ...

5

Judging by the time of your post, I would probably say the star you saw was Aldebaran as it was very close to the moon on that night. It's the brightest star in the constellation Taurus so is easily seen close to the moon, where the moonlight washes out the dimmer stars around it. It is normally seen red/orange to the naked eye but atmospheric interference ...

5

Is it just for the spoiler? My guess is that the 3 suns aligned create a sufficient gravity field for the outermost parts - less well attached - of the planet (atmosphere, people, buildings) to be in the Roche Limit of the system, causing them to be sucked out spiraling inwards. Here is a funny vid I found to illustrate: https://www.youtube.com/watch?v=...

5

The initial star formation regions were regions that have a high enough mass density to form a star. The density of the early universe was not constant at different locations. Some regions had high enough density to form a star, and some didn't. When a star forms it draws in matter from a large distance away. This forms an accretion disk and leaves a ...

5

As jmh has answered, stars naturally form at large distances from each other. To add to the answer, what is the reason for this particular distance scale? If we imagine a very large, homogeneous cloud of gas it will be unstable to gravitational collapse over the Jeans length scale $\lambda_J=c_s/\sqrt{G\rho}$ where $c_s$ is the speed of sound in the gas, $G$ ...

5

The ratio $T/\Omega$ tells you about the acceleration of the system - or more specifically, the second derivative of its moment of inertia - it does not tell you about the velocity. If the system collapses because it has $T/\Omega<0.5$, then when it reaches $T/\Omega=0.5$ it stops accelerating. That doesn't mean it stops collapsing. It overshoots in the ...

5

The mass of a star is determined by its gravitational effects, not by summing up the components that give rise to that gravity. If we were to do the latter, then we would have to include all the mass and energy in the star. This would comprise of the rest mass of all particles, the kinetic energy of the particles and the energy of massless particles like ...

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