Deriving the radius

The heliopause is a place of equilibrium, where the ram pressure from the solar wind is equal to the pressure of the interstellar medium (ISM). There are a number of sources of pressure in the ISM, but thermal pressure is the main one.¹ The ram pressure from a wind with terminal velocity $v_{\mathrm{\infty }}$ at a distance $r$ from the a star is $$P_R=n_{\ast }(r)m{v}_{\mathrm{\infty }}^2$$ where $n_{\ast }(r)$ is the number density of the wind, and $m$ is the mean mass of the particles at the heliopause. It turns out that, by mass conservation, $n_{\ast }(r)\propto \dot{M}{v}_{\mathrm{\infty }}^{-1}{r}^{-2}$, where $\dot{M}$ is the mass-loss rate.² The thermal pressure of the ISM is $$P_T=n_I{k}_{B}T$$ where $k_B$ is Boltzmann's constant and $T$ is the temperature. $n_I$ is the number density of the interstellar medium. For Sun-like stars, $v\sim 400$ km/s is reasonable. $n_{\odot }$ and $n_I$ are about 5 particles per cubic centimeter and 0.01 particles per cubic centimeter, respectively, and $T\sim {10}^5$ K. Assuming the wind is largely hydrogen at this point, if we set the two equal, we get $$\begin{array}{rl}{r}_{\text{crit}}& ={\left(\frac{n_{\text{crit}}m{v}_{\mathrm{\infty }}^2}{n_I{k}_{B}T}\right)}^{1/2}\text{ AU}\\ & \approx 311{\left(\frac{n_{\text{crit}}}{5{\text{ cm}}^{-3}}\right)}^{1/2}\left(\frac{v_{\mathrm{\infty }}}{400\text{ km/s}}\right)\text{ AU}\\ & =311{\left(\frac{\dot{M}}{{10}^{-14}{M}_{\odot }{\text{ yr}}^{-1}}\right)}^{1/2}{\left(\frac{v_{\mathrm{\infty }}}{400\text{ km/s}}\right)}^{1/2}\text{AU}\end{array}$$ where $n_{\text{crit}}\equiv n_{\ast }(r_{\text{crit}})$. I've used scaling relations such that for the Sun, $r_{\text{crit}}$ is 311 AU.

Important factors

Some things to note:

• Stellar winds don't all have the same composition, but hydrogen is, by and large, the major component, and the one important factor when it comes to calculating $m$.
• The most important star-dependent variables are $\dot{M}$ and $v_{\mathrm{\infty }}$. Note that $r_{\text{crit}}\propto {\dot{M}}^{1/2}{v}_{\mathrm{\infty }}^{1/2}$. For massive O- and B- stars, winds can have speeds of $\sim 2000$ km/s, or more; I think some of the strongest are about $\sim 3000$ km/s. This could mean heliopauses of thousands of AU. HD 93129a is a good example, with $v_{\mathrm{\infty }}$ of about $\sim 3000$ km/s.
• I can try to pull some numbers for Proxima Centauri, but I'll point out that red dwarfs usually don't have strong stellar winds. The interest in the wind of Proxima Centauri is really because Proxima Centauri b orbits close enough to the star that stellar activity - especially flares - could cause severe problems for life.

Star types

For Sun-like stars, $\dot{M}\sim {10}^{-14}{M}_{\odot }$ per year is reasonable, and so you'd get heliopauses of a few hundred AU. For O- and B- type main sequence stars, I'd expect $\dot{M}\sim {10}^{-6}{M}_{\odot }$ per year, which is much higher. At the other end of the spectrum - those red dwarfs that are relatively quiet - we'd see maybe $\dot{M}\sim {10}^{-15}{M}_{\odot }$ per year. A- and F- main sequence stars might have mass-loss rates a bit higher than the Sun, and perhaps larger terminal velocities.

Off the main sequence, things get more complicated. Red giants - especially asymptotic branch stars, near the end of their lives - have large mass-loss rates that arise via different mechanisms involving dust. These are cool but bright stars; consider $\dot{M}\sim {10}^{-8}{M}_{\odot }$ and reasonable fast winds, though not nearly as fast as the hot, massive O-type stars. Additionally, more exotic cases like Be stars that are undergoing mass loss may have larger mass-loss rates. Finally, the young T Tauri stars have mass-loss rates similar to red giants - maybe an order of magnitude or so lower - but their winds are slower than the Sun's.

Specific cases

I looked around and found instances where $\dot{M}$ and $v_{\mathrm{\infty }}$ have been observed, and made an estimate of $r_{\text{crit}}$: $$\begin{array}{|cccccc|}\hline \text{Star}& \text{Stellar type}& \text{Mass }(M_{\odot })& \dot{M}(M_{\odot }{\text{ yr}}^{-1})& v_{\mathrm{\infty }}(\text{km/s})& r_{\text{crit}}(\text{AU})\\ {\text{HD 93129Aa}}^1& \text{O2}& 95& 2\times {10}^{-5}& 3200& 3.93\times {10}^7\\ \tau {\text{ Sco}}^2& \text{B0}& 20& 3.1\times {10}^{-8}& 2400& 1.32\times {10}^6\\ \sigma {\text{ Ori E}}^3& \text{B2}& 8.9& 2.4\times {10}^{-9}& 1460& 2.91\times {10}^5\\ \alpha {\text{ Col}}^2& \text{B7}& 3.7& 3\times {10}^{-12}& 1250& 9.52\times {10}^3\\ {\text{Deneb}}^4& \text{A2}& 20& {10}^{-6}& 225& 2.3\times {10}^6\\ \text{Sun}& \text{G2}& 1& {10}^{-14}& 400& 311\\ {\text{Proxima Centauri}}^5& \text{M6}& 0.12& {10}^{-13}& 550& 1332\\ \hline\end{array}$$ _{¹Cohen et al. (2011)
²Cohen et al. (1997)
³Krtička et al. (2006)
⁴Aufdenberg et al. (2002)
⁵Wargelin & Drake (2002)}

Now, HD 93129Aa and Deneb are supergiants, so they're off the main sequence, but their properties here shouldn't be too far off from main sequence stars of the same spectral type. Deneb's mass-loss rate is maybe a bit high in comparison to main sequence A stars. Also, I'm slightly skeptical of the value for HD 93129Aa's heliopause, so it's possible that other factors play a role - for instance, thermal pressure could indeed be important in its hot wind. Additionally, some M dwarfs have higher stellar winds and mass-loss rates because of flares and other activity.

¹ We can disregard ram pressure, as the ISM is, by and large, slow-moving. Likewise, magnetic fields are typically not important. Similarly, we can neglect thermal pressure in the stellar wind; even though winds may have temperatures of several million K, ram pressure is more important.
² Specifically, the density ${\rho }_{\odot }(r)$ (related to number density by ${\rho }_{\odot }(r)=m{n}_{\odot }(r)$) is given by $${\rho }_{\odot }(r)=\frac{\dot{M}}{4\pi r^{2}v(r)}$$ where $\dot{M}$ is the mass loss rate and $$v(r)=v_{\mathrm{\infty }}{(1-\frac{R_{\ast }}{r})}^{\beta }$$ with $R_{\ast }$ being the radius of the star. We typically assume $\beta \approx 1$, but at $r\gg R_{\ast }$, we can say that $v(r)\approx v_{\mathrm{\infty }}$.

posted almost 7 years ago

HDE 226868‭

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