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I presume I'm missing something here but from looking at charts indicating the value of Mach 1 at different altitudes it seemed to reduce the higher in the atmosphere you went which I presume is related to reduced temperature and not pressure related which to me seemed counter intuitive. Thanks, hopefully you can clear this up

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    $\begingroup$ While there are people who understand physics well around here, it would probably make more sense to ask this over on physics. $\endgroup$
    – Jan Hudec
    Commented Jul 5, 2017 at 17:11
  • $\begingroup$ Sound speed depends both on pressure and density which cancel each other in an ideal gas ($c=\sqrt{\gamma \frac{p}{\rho} } $). But actually pressure has a small impact on real air. See Wikipedia. $\endgroup$
    – mins
    Commented Jul 5, 2017 at 17:16

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It may be helpful to remember the ideal gas law here: $PV=Nk_BT$ (the physics version). If we divide both sides by $N$ we have a $V/N$ on the left which is just a $1/\rho$ where $\rho$ is the number density. Thus:

$\frac{P}{\rho}=k_BT$

In an ideal gas, to which air is close enough, the speed of sound is given by:

$c_s = \sqrt{\gamma \frac{P}{\rho}}$

Throwing in our first results, we see that:

$c_s = \sqrt{\gamma \frac{P}{\rho}} = \sqrt{\gamma k_BT}$

In other words, the speed of sound reduces with either pressure or temperature because those two are in turn connected by the ideal gas law. It's basically two different ways of looking at the same question.

Side note: throwing in the effect of density is another question all together since $c_s$ increases with an increase in temperature or pressure but decreases with an increase in density. The question then becomes whether the decrease due to the drop in pressure overcomes the increase due to the drop in density at higher altitude.

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  • $\begingroup$ The side note is quite relevant. You show that speed of sound is a constant times temperature, but without knowing more about how pressure varies with density, it cannot be stated that speed of sound is a function of pressure. $\endgroup$
    – Koyovis
    Commented Jul 7, 2017 at 2:35

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