Speed of sound at the Sun surface [closed]

The gas velocity at the Sun's surface is subsonic and At 1 AU from the sun, the solar wind is supersonic. Like Parker wind solution.

Why the velocity at the Sun's surface is subsonic?

(In the isothermal condition but I think also in general)

closed as unclear what you're asking by Kyle Kanos, stafusa, M. Enns, glS, AccidentalFourierTransformJul 24 '18 at 15:59

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• Can you add a source, a reasoning...? – QuirkyTurtle98 Jul 9 '18 at 17:13
• @QuirkyTurtle98, I've heard this in a seminar on astronomy. However, this compatible with Parker wind solution. Actually, I'm searching for the reason! – Rahaa Jul 10 '18 at 10:45
• Perhaps this will help: physics.stackexchange.com/a/179057/59023 – honeste_vivere Jul 19 '18 at 15:49

The speed of sound is simply $\sqrt{5P/3\rho }\sim \sqrt{T}$, and the polytropic rule says that $\rho \sim {{T}^{3/2}}$.
The speed of convection currents cannot be calculated with any precision, but it may be estimated via glorified dimensional analysis. Naive dimensional analysis is ambiguous because too many parameters seem relevant: the luminosity (Q), radius (R), density $(\rho )$, coefficient of thermal expansion (a), specific heat capacity $({{C}_{P}})$, and acceleration of gravity (g). However, a detailed look at convection narrows down the relevant combination. When a blob of plasma absorbs heat, it becomes buoyant, and the heat is converted to kinetic energy as the blob rises through a distance $Z={{C}_{P}}/ag=T{{C}_{P}}/g$. Kolmogorov’s theory of turbulence identifies power per unit mass $(Q/Z{{R}^{2}}\rho )$ as the key combination. The meaningful combination with the units of speed is $\sqrt{Q/ZR\rho }\sim {{T}^{-5/6}}$. According to this argument, convection gets faster and overtakes the speed of sound as you approach the surface outbound.