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 \(O_2\),
\(\cfrac{1}{2}m_{\rho}v^{ 2 }_{ { max } }=E_{input/proton}=\cfrac{1}{2}c^{ 2 }*3.4354*1.429*\cfrac{1}{16}\)
\(v_{\small{O_2\,boom}}=0.3068\,ms^{-1}=1.10\,kmh^{-1}\)
