3
$\begingroup$

I need someone to give me a way to figure out how fast the solar wind from a star with a known mass travels, as well as how much solar wind the star releases in a given amount of time. I hope that it helps me determine how magnetic a planet needs to be when orbiting a star that’s not the sun to not get its atmosphere sheared off by the solar wind.

$\endgroup$
2
  • $\begingroup$ @ARogueAnt. Thanks. $\endgroup$ Commented Apr 23, 2021 at 22:15
  • $\begingroup$ Astronomy Stack Exchange. (And it all depends on the atmosphere, doesn't it? For example, Venus has a massive atmosphere, although it doesn't have any magnetic field to speak of, and being closer to the Sun it experiences a much stronger solar wind than Earth. True, there is no water in Venus's atmosphere, due to photodissociation and hydrogen being blown off.) $\endgroup$
    – AlexP
    Commented Apr 23, 2021 at 22:21

1 Answer 1

6
$\begingroup$

This is a difficult (but interesting!) question to answer, and strongly depends on the star in question. In general, hot, massive stars have higher mass-loss rates and higher terminal velocities then lower-mass stars, but that may change as a star evolves. For instance, stars like the Sun eventually evolve off the main sequence into asymptotic giant branch stars, which may lose up to $10^{-4}M_{\odot}$ per year, which is about ten orders of magnitude more than on the main sequence. Other difficulties arise because there's no universally applicable wind mechanism; massive stars tend to have radiation-driven winds, while less-massive stars may have thermally-driven winds arising from pressure gradients.

Here's a table of the wind properties of a set of stars, which I pulled from an earlier answer of mine. HD 93129Aa and Deneb are supergiants, while the rest are main sequence stars. I've sorted them by spectral type, but if you sort by mass, you'll notice that in general, lower-mass stars have lower mass-loss rates $\dot{M}$ and lower terminal velocities $v_{\infty}$.

$$ \begin{array}{|c|c|c|c|c|}\hline \text{Star} & \text{Stellar type} & \text{Mass }(M_{\odot}) & \dot{M}(M_{\odot}\text{ yr}^{-1}) & v_{\infty}(\text{km/s})\\\hline \text{HD 93129Aa}^1 & \text{O2} & 95 & 2\times10^{-5} & 3200\\\hline \tau\text{ Sco}^2 & \text{B0} & 20 & 3.1\times10^{-8} & 2400\\\hline \sigma\text{ Ori E}^3 & \text{B2} & 8.9 & 2.4\times10^{-9} & 1460\\\hline \alpha\text{ Col}^2 & \text{B7} & 3.7 & 3\times10^{-12} & 1250\\\hline \text{Deneb}^4 & \text{A2} & 20 & 10^{-6} & 225\\\hline \text{Sun} & \text{G2} & 1 & 10^{-14} & 400\\\hline \text{Proxima Centauri}^5 & \text{M6} & 0.12 & 10^{-13} & 550\\\hline \end{array} $$ 1Cohen et al. (2011), 2Cohen et al. (1997), 3Krtička et al. (2006), 4Aufdenberg et al. (2002), 5Wargelin & Drake (2002)

As you might expect, the speed and density of the wind aren't constant at all points around a star. Spherical symmetry is often a good choice (although rotation and magnetic fields can complicate this close to the star's surface). With a couple of additional assumptions, we get the following expressions for the density of a star's wind $\rho$ and the wind speed $v$ at a radius $r$: $$\rho(r)=\frac{\dot{M}}{4\pi r^2v(r)},\quad v(r)=v_{\infty}\left(1-\frac{R_*}{r}\right)^{\beta}$$ The latter is something of an assumption, but it does work reasonably well, particularly for massive stars, where we usually assume $\beta\approx1$.

I assume you're looking at a lower mass Sun-like star for the sake of habitability, so a mass-loss rate of $\sim10^{-13}M_{\odot}$ per year and a terminal velocity in the range of 400-500 km/s would be reasonable. You can then calculate the density and speed at the orbital radius of the planet.

$\endgroup$
1
  • $\begingroup$ Thanks for giving me an answer. I’ll be using your formulas later. $\endgroup$ Commented Apr 24, 2021 at 1:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .