A Small Boom

From the post "Sonic Boom" dated 14 Oct 2017, for water (not steam),

$\cfrac{1}{2}m_{\rho}v^{ 2 }_{ { max } }=E_{input/proton}=\cfrac{1}{2}c^{ 2 }*3.4354*1000*\cfrac{1}{10}$

If we replace water with the most abundant element in air, nitrogen, $N_2$,

$\cfrac{1}{2}m_{\rho}v^{ 2 }_{ { max } }=E_{input/proton}=\cfrac{1}{2}c^{ 2 }*3.4354*1.165*\cfrac{1}{14}$

which has a density of $1.165\,\, kgm^{-3}$ with a total of 14 particles (adding Atomic Numbers) in the molecule.  We have

$E_{input/proton}=\cfrac{1}{2}(one\,\,particle)*c^{ 2 }*(one\,\,unit\,\,volume)*0.286$

we have a low boom at $0.286\,\, ms^{-1}$ or $1.03\,\, kmh^{-1}$

what is the significance of this speed?

This calculations however is based on resonance, not of a minimum threshold, beyond $343.5\,ms^{-1}$ or $1.03\,\, kmh^{-1}$ resistance due to the booms should decrease rapidly.  The $Q$ factor of the underlying resonance phenomenon determines the rise and fall of resistance around the boom values.

The nitrogen boom can be observed as a small boat accelerates to high speed.  This boom is experienced as a bump in the wave as the bow lifts due to the boom.  But as speed increases the resistance decays quickly.  If the boat remains at nitrogen boom speed, $1.03\,\, kmh^{-1}$, it will flip over, bow backwards.

If we are able to obtain the Q profiles as speed varies around $343.5\,\,ms^{-1}$ or $1.03\,\, kmh^{-1}$, the Q factors are indicative of the underlying resonance phenomena.

Good night.

Note:  We also have for $O_2$,

$\cfrac{1}{2}m_{\rho}v^{ 2 }_{ { max } }=E_{input/proton}=\cfrac{1}{2}c^{ 2 }*3.4354*1.429*\cfrac{1}{16}$

$v_{\small{O_2\,boom}}=0.3068\,ms^{-1}=1.10\,kmh^{-1}$