# Why does the speed of sound decrease at high altitudes although the air density decreases?

I understand that the speed of sound is inversely proportional to the density of the medium as shown here and as answered for this question.

The problem now is that the speed of sound in air actually decreases with altitude although the density of the air decreases. This is shown here and here.

I understand that the speed of sound also depends on the elasticity, but I'm not sure how this can change for air.

So what is actually happening? How can the speed of sound decrease although the density has also decreased?

• Good question! A small comment: When linking to graphs as sources, it would help to link the article where the image is from - that helps the reader to understand what the graph means and how the values were determined.
– JiK
Sep 13 '19 at 19:40

Wikipedia gives a pretty much straightforward answer. In an ideal gas, the speed of sound depends only on the temperature:

$$v = \sqrt{\frac{\gamma \cdot k \cdot T}{m}}$$

So it neither decreases, nor increases with altitude, but just follows air temperature as can be seen in this graph:

• Comments are not for extended discussion; this conversation has been moved to chat.
– rob
Sep 13 '19 at 14:18
• "The speed of sound in an ideal gas depends only on its temperature and composition. The speed has a weak dependence on frequency and pressure in ordinary air, deviating slightly from ideal behavior." - Wikipedia Mar 26 at 15:10
• @ViniciusACP According to ideal gas law, temperature maps to pressure directly. Agree, that there's a sound wave dispersion, but for sake of reason these can be ignored, as long as graph shows strong correlation only between $T$ and $v$. Mar 26 at 20:59

The speed of sound in a gas is given by $$\sqrt{ \dfrac {\gamma \,P}{\rho}}= \sqrt{\gamma \, R \, T}$$ where the temperature, $$T$$, is in kelvin, $$\gamma$$ is the ratio of the specific heat capacities of a gas at constant pressure and constant volume and $$R$$ is the specific gas constant.

With increasing altitude there is a decrease in the density but also a decrease in the pressure, but not at the same rate because there is also a change in the temperature.

As the altitude increases, the temperature decreases and so does the speed of sound and then when the temperature increases so does the speed.

Source - Engineering Toolbox

• Have you built this table? Very good and my best complimens. Sep 13 '19 at 21:27
• On the left-hand side of your equation, $\rho$ is mass density, not molar density. This means that the right-hand side is incorrect as well — it's dimensionally inconsistent. I believe that the correct expression would be $\sqrt{ \gamma R T/m_\text{mol}}$, where $m_\text{mol}$ is the molar mass. Sep 13 '19 at 21:36
• @MichaelSeifert I understand what you have written and the apparent error is due to the fact that I did not define my gas constant $R$ as the specific gas constant with units of $\rm J\, kg^{-1} \, K^{-1}$. Sep 13 '19 at 21:48
• @Sebastiano The table comes from the Engineering Toolbox website. I have also put the source of the table in my answer. Sep 16 '19 at 10:10

Sound is simply a compression wave. The velocity of the wave is inversely related to the square root of fluid pressure, and directly proportional to the materiel’s Young modulus. A lower density means a lower pressure, which increases the wave velocity as you noticed.

Traveling through a compressible medium such as air, the simple equation for the velocity (v) of a compression wave is the square root of the Young’s modulus (B) divided by pressure (p)

$$v=\sqrt{\frac{B}{p}}.$$

• $B$ is the bulk modulus not Young’s modulus. Sep 13 '19 at 8:13
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}