# Why does air density drop when speed nears the Mach 1?

In many questions and answers about high speed aerodynamics here it is said that as speed approaches the speed of sound, there is also a decrease in air density.

Can someone explain why?.

• Can you add a link to at least one such question?
– Jim
Aug 26, 2022 at 14:38

We start with the Laplace equation for the speed of sound: $$a = \sqrt{\frac{\gamma \cdot p}{\rho}}$$ and differentiate and square it: $$a^2 = \frac{\delta p}{\delta\rho}$$ Now we can write: $$\delta p = \delta\rho \cdot a^2$$ Next to be replaced is the pressure differential $$\delta p$$, using the equation for the conservation of momentum in its differentiated form: $$\delta p = -\rho \cdot v \cdot\delta v$$ $$\rightarrow\:-\frac{\delta v}{v} = \frac{\delta\rho}{\rho} \cdot \left(\frac{a}{v}\right)^2$$ which can be written as $$-\frac{\delta v}{v} \cdot Ma^2 = \frac{\delta\rho}{\rho}$$ Interpretation: At small Mach numbers, changes in speed cause negligible changes in density, but as Mach approaches unity, both are of similar magnitude. With $$Ma>>1$$, changes in density will become dominant. A speed increase is always coupled to a decrease in density.