I'm wondering if different stars produce wind at different speeds, and if so, what factors determine that speed.

Per this answer, our star produces wind with a speed between ~250–820 km/s. It seems safe to assume that the wind speed for each star is variable.

This other answer gives the Reimers mass-loss formula, which accounts for loss via radiation and wind, but states that one term must be measured empirically. (And, in any case, I'm interested in wind speed, not mass carried away.)

Wikipedia says we don't have a very good understanding of how the solar wind is accelerated, which suggests it may not be possible to answer this question.

Still, as a layman, it seems like it should be possible to make general predictions about a star's wind, given a few facts about that star, namely: mass, angular velocity, and age.

For example, would a blue star have faster wind because of its higher temperature or (I assume) stronger magnetic activity? Or would it be slower because of its greater gravity? Or would it depend substantially on the star's rotation? Or would all stars have similar wind speeds because of facts about idealized gasses (I'm thinking of Maxwellian distribution)?


2 Answers 2


The answer depends very much on what type of star you are talking about and what the driving mechanism of the wind is. Broadly we can split stars up into low-mass main-sequence stars (say $<2 M_\odot$), where the wind is probably thermally driven by gas pressure gradients, high-mass main-sequence and evolved stars, where the wind is driven by radiation pressure, and red giants where the wind is also driven by radiation pressure, but the opacity source is dust.

For low-mass main sequence stars, there are no observations to make an empirical relationship and there are major theoretical uncertainties in estimating a wind velocity. This is discussed by Johnstone et al. (2015a) who find the wind velocity is (at 1 au) $$v = 73.39 + 221.14T_0 - 11.28T_0^2 +0.28T_0^3\ , $$ where $T_0$ is the base temperature of the wind. The main uncertainty is whether the base temperature of the wind is determined by the magnetic fields present (since the corona is ultimately heated by magnetic fields), or whether it is directly connected to the stellar escape velocity. If the former, then the wind speed will be connected with the rotation rate of the star and its spectral type in a very non-linear way. Some plots are presented by Johnstone et al. (2015b) and I reproduce one of these below. The rotation dependence is manifested as an age dependence since these stars spin down as a result of their magnetised winds. This in turn reduces the dynamo-induced magnetic activity and reduces their coronal temperatures.

Johnstone et al. 2015 wind speed

Hot, early-type O and B stars have winds driven by radiation pressure. The theory of "line-driven" winds is complex and cannot be simply summarised in a SE answer. The terminal wind velocity is usually 2-3 times the escape speed at the star's photosphere. But there is a complex detailed dependence of this factor on spectral type and hence luminosity (or mass). The terminal wind speeds are in the range 1000-3000 km/s for B $\rightarrow$ O-stars. This relationship of a few times the escape velocity works reasonably well for Wolf-Rayet and other hot, evolved stars too.

The winds from cool giants are radiatively driven, but this time the opacity source is dust. For these stars, the terminal wind velocity is given approximately by $$ v_{\infty} \simeq \frac{v_{\rm esc}^2}{1000\ {\rm km/s}}\ , $$ which is lower than the escape velocity. Roughly 10 km/s for M-type supergiants, up to 70 km/s for K-type giants. Higher velocities are not found because the dust disintegrates by sputtering once accelerated to these speeds.


All stars, and also planets, lose mass because of atoms escaping from the atmosphere that are faster than the gravitational escape velocity. This suggests already that gravity i.e. the mass of stars is the crucial factor here. The gravitational potential energy of an atom with mass m near the surface of a star with mass M and radius R is of the order

$$E_{pot} =-\frac{GMm}{R}$$

where $G$ is the gravitational constant.

and according to the virial theorem, the associated kinetic energy is


which corresponds to a temperature


where $k$ is the Boltzmann constant.

If you evaluate this for the Sun, you get

$$T=1.1\cdot 10^7 K$$

which corresponds to a thermal speed of hydrogen of

$$v=\sqrt{\frac{2kT}{m}} \approx 400 km/sec$$

This is essentially the solar wind speed we observe.

Given the fact that the radius $R$ in the denominator in the first equation increases with mass as well (depending on the mass-radius relationship) and the square root in the last equation, the dependence of the stellar wind speed on the mass of the star should be fairly weak though.

Of course this would only be an average speed. The fact that we are dealing with charged particles here introduces all kinds of electromagnetic effects that can change the observed speed quite drastically in certain situations.

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    $\begingroup$ I think it would be good to specify that this explanation only may explain the very small mass loss from low-mass stars (e.g. the Sun). The much higher mass loss from high-mass stars is caused by radiation pressure, not thermal escape. $\endgroup$
    – pela
    Commented May 2, 2022 at 9:22
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    $\begingroup$ A great textbook for learning about stellar winds is by Lamers and Cassinelli, and here is an abridged version for free with many chapters. They have a great intro on the winds of high mass stars. ui.adsabs.harvard.edu/abs/1999isw..book.....L/abstract $\endgroup$ Commented May 2, 2022 at 10:36
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    $\begingroup$ @pela The OP was asking about the solar wind speed not the mass loss. The two are not necessarily related. If the Sun had a thinner photosphere (i.e. fewer of the particles with 'virial' energy would be stopped due to inelastic collisions with neutrals) the mass loss of the Sun would be much higher but the energy of the solar wind would still be the same, given essentially by the gravitational potential energy i.e. the mass. $\endgroup$
    – Thomas
    Commented May 2, 2022 at 11:43
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    $\begingroup$ @CarlWitthoft A rotation shouldn't actually make a difference to the consideration above. The virial theorem relates the gravitational potential energy to the total kinetic energy. If the latter contains rotational energy as well, then the thermal energy must be correspondingly less for the whole system to be in equilibrium. Added together, the total kinetic energy and thus the escape speed will still be the same as for a non-rotating star (which the Sun with about 2 km/sec surface speed pretty much is). $\endgroup$
    – Thomas
    Commented May 2, 2022 at 13:58
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    $\begingroup$ The OP asks about how to calculate a stellar wind speed based on the parameters of the star. This answer uses a virial calculation that assumes the gas at the surface of the Sun is in a virial equilibrium. The virial theorem cannot be applied to individual components of a system like this. Without non-radiative magnetic heating, the region above the photosphere would be cooler than the photosphere and the Sun's corona is in any case much cooler than $10^7$K on average. $\endgroup$
    – ProfRob
    Commented May 2, 2022 at 14:20

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