# Tag Info

42

Models for the future behaviour of the Sun do vary, mainly as a result of uncertainty of mass loss during the red giant (H shell burning) and asymptotic red giant (H+He shell burning) phases. A highly cited paper by Schroeder & Smith 2008 claims that the Sun will reach its maximum size of about $256 R_{\odot}$ (1.18 au) at the very tip of the red giant ...

17

Arcturus is a RGB star, probably fairly similar how the sun will look when it becomes a red giant. Arcturus is slightly more massive than the sun ($m_{\rm Arc}=1.08 m_{\odot}$), but the main difference is the lower metallicity of $[Fe/H]\approx-0.5$. This low metallicity reduces the opacity in the stellar radiative zone (which fills a significant portion of ...

11

(This is somewhat simplified but I hope it gets the idea across.) The reactions stop in the core because it runs out of fuel. During the main sequence, the star is supported by the fusion of hydrogen into helium. Eventually, the hydrogen runs out at the centre, so hydrogen fusion is no longer possible there. Why doesn't it start fusing helium into carbon ...

11

It will become a planetary nebula like e.g. the Cat's Eye nebula that was formed by the death of a star with about the same mass as the Sun ~ $1\ M_\odot$:                              Composite image using optical ...

10

Indeed conservation of angular momentum dictates that in a single star like the the Sun, rotation should be much slower when it becomes a red giant. This is because at the present time the Sun does not rotate at vastly different rates with depth, thus when it expands, the moment of inertia increases drastically and convection in the outer envelope will ...

10

In my mind, none of these explanations really cover the actual reason that red giants expand. Indeed, this subject seems like an area where people just make up anything that sounds plausible, but it's often quite wrong (Fraser Cain mentions both light pressure and a higher volume in the fusing shell, but light pressure plays no role at all, and the volume ...

8

There are some answers in this lecture and this one although even there the author admits that the whole story is complicated, and not perfectly known. It seems that shell burning and contraction of the no-longer-burning core of the star causes some expansion of the outer layers (this is described as the "mirror principle" and partly explained in lecture ...

8

The answer is yes, there are many authors who do take mass loss from stars into account when trying to work out the fate of their planetary systems. Examples include Schroder & Smith 2008; Adams et al. (2013); Adams & Bloch (2013). It is thought that (e.g. Kalirai et al. 2008) that the Sun will lose half its mass by the time it ends up as a white ...

7

There's a nice description here. Remember that a star is made of gas (well plasma if you want to be picky), so it doesn't have a fixed volume. Once fusion starts a star will expand until it reaches a size where it can balance the amount of energy being produced by fusion with the amount radiated away from the surface. If it's too small, it will heat up, ...

7

I'll try to add a bit of context to the Wikipedia article, though the major references are all available there. The article covers things like the potential formation mechanisms but I guess doesn't really put them into scientific perspective. Consider, for a moment, a red giant of one solar mass, far up the red giant branch. It has a deep convective ...

7

The answer is of order 1 million years to cool from a standard end of He burning temperature of just over $10^8$ K to the top end of the white dwarf temperature range you give in your question. The details would depend exactly on the mass and composition of the white dwarf and there are also some theoretical uncertainties in neutrino cooling rates. The ...

6

The difference is that your analysis is assuming that the albedo stays fixed, so the surface temperature simply scales like luminosity to the 1/4 power. The Wiki entry is including feedback from the greenhouse effect, which will tend to further increase the surface temperature. Note that an analysis that just looks at solar irradiation would get way too ...

6

The real reason stars bloat into red giants is not due to opacity changes, it is due to the creation of a degenerate core of non-fusing helium at the center. This degenerate core has a strong gravity, which dictates a high temperature to the hydrogen fusing shell that sits atop it. This is quite important, because when a star is undergoing core fusion like ...

6

Rayleigh scattering happens at all wavelengths, but the scattering cross section goes as $\lambda^{-4}$. On Earth, the atmospheric optical depth to Rayleigh scattering is very small at red wavelengths, so hardly any red light is scattered, even at sunset when the Sun is viewed through a thick atmospheric layer. On the contrary, there is sufficient optical ...

6

Yes, you can see one tonight. Arctaurus is a red giant star with a mass of about 1.1 times the solar mass, so rather similar to the sun. It currently has a spectral type of K0 III. It is ascending the red giant branch, so it's luminosity and spectrum are not stable in the longer term. The sun will pass through this phase, and following a hydrogen flash ...

5

The size of star in equilibrium is a balance of forces, the pressure produced by the hot plasma, heated by the nuclear reactions in the core, balanced by gravity. Fusion rates are strongly affected by temperature. Increase the temperature a little and you get a lot more energy coming out. As the core runs out of hydrogen, it begins to collapse and heat up, ...

5

Yes, the time is very long, but exactly how long depends on when you "start the clock" for the expansion to a red giant. If you take the end of the "main sequence" (when the core hydrogen runs out) as the start of the expansion phase, it's more like a billion years to complete the process of becoming a maximally expanded red giant. But the answer that gets ...

5

The Sun will not become a red giant for approximately 7 billion years, so there is little chance that Proxima Cen will be anywhere near it. [EDIT: To answer a comment, the current helocentric radial velocity of Proxima Cen is -22 km/s. The Sun's escape velocity at the distance of Proxima Cen is about 0.1 km/s. So the Sun and Proxima Cen are totally unbound. ...

5

Low mass stars like the Sun do become very large prior to He ignition in the core. The exact value depends a bit on models for mass loss from the extended atmosphere (e.g. Guo et al. 2016), but estimates of 250 times its current size are possible (Schroeder & Smith 2008; Spiegel & Madhusudhan 2012). At this radius both Mercury and Venus are engulfed. ...

4

The straightforward definition is in terms of where a star lies on its evolutionary track in the HR diagram (see below). The subgiant branch stars are those which have exhausted their hydrogen core and which are burning hydrogen in a shell but their He cores have not begun to contract significantly. The distinct upturn in luminosity marks the beginning of ...

4

The AAS Nova article cites Rawls et al. 2016, who analyze light curves and spectra of the eclipsing binary KIC 9246715 to estimate its physical properties. Besides the stellar masses and radii in the article, their Table 2 lists orbital semimajor axis a = 211 R☉ = 147 million km = 0.98 au, and orbital eccentricity e = 0.356. During the 171-day period, ...

4

The year length depends on the distance between the planet's centre & the Sun's centre, not the Sun's surface. So if the Sun merely expands, Earth at 1 AU will still take a standard year to perform 1 orbit. However, when the Sun becomes a red giant, it won't just expand. As Wikipedia mentions, red giants shed a considerable amount of mass in the form of ...

3

I will replace the previous answer to focus on the "subgiant" branch prior to red giant, rather than the pre-main-sequence or the "horizontal branch" of core helium fusion. Those are other times that the luminosity is constant, but this question is about the subgiant branch, which I missed before. The reason the luminosity is nearly constant on the ...

3

The destiny of a star basically depends upon its mass. All its activities variety depends upon its mass. If a star's core has a mass that is below the Chandraseckhar limit ($M\sim1.4M_{sun}$), then is destined to die as a white dwarf (or, actually, as a black dwarf in the end). The composition of the white dwarf, also depends upon the original mass of the ...

3

Where does the evolution actually stop? Is the Helium Core Flash the last thing a Sun-like stars experiences or does it in fact follow the rest of the tractory? Stars with masses below about 2 solar masses undergo a helium flash but their evolution does continue afterwards, as shown in the second diagram (Fig. 13.4). So no, the core flash is not the last ...

3

Higher mass stars will have shorter lives. Even though they have more fuel for nucleosynthesis, they burn this fuel much quicker than lower mass stars. Generally, you should think of "red-giant" as an evolutionary phase and not a particular type of star. Looking at the Scheller et al. (Aston. Astrophys. Suppl., 96, 269, 1992) data given in one of ...

2

The distance to betelgeuse is poorly known, so we don't actually know how bright it is with much accuracy measurements of its parallax by satellite give a distance of 197 parsecs +/- 45 parsecs (1 parsec is 3.26 light years). The absolute magnitude (the brightness if it were 10 parsecs distant) is estimated to be -5.85. This is based on both the distance and ...

2

Given a distance of $d_1 = 640~\mathrm{ly}$ for Betelgeuze and $d_2 = 8.6~\mathrm{ly}$ for Sirius the difference in the apparent magnitude for Betelgeuze at distances $d_1$ and $d_2$ that is given by $$m_1 - m_2 = 5~\mathrm{mag} \cdot log_{10}(\frac{d_1}{d_2})$$ is $9.4~\mathrm{mag}$. With magnitude $m_1 = 0.45~\mathrm{mag}$ we get $m_2 = -8.9~\mathrm{mag}$...

2

For a more fundamental understanding, it is helpful to realize the difficulties of fusing He-4 into C-12. This is called the Triple-Alpha process. When two He-4 nuclei (alpha particles) have sufficient energy to overcome to the Coulomb barrier and have their cross sections align, it produces Be-8. The Be-8 nucleus is so unstable (due to it being ...

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