Exercises — Shock waves — Mach number, Mach cone — - CRITICAL for rockets -

WHAT we do: form the ratio $M$ — the only thing that decides the regime. $M=\frac{v_{\text{obj}}}{v_s}=\frac{270}{340}=0.794$ WHY: raw speed alone can't tell us — 270 m/s is fast, but the question is did it beat its own ripples? Since $M<1$, it did not. Answer: $M\approx 0.79$ → subsonic. Because $\sin \theta =1/M=1.26>1$ has no real angle, no Mach cone exists. The object stays inside its own expanding wavefronts.

WHAT/WHY: invert the one formula. $\sin 90^{\circ }=1$, so $\sin \theta =\frac{1}{M}=1\Rightarrow M=1.$ WHAT IT LOOKS LIKE: the cone has opened all the way out to a flat plane perpendicular to the flight path — the wavefronts pile into a single wall directly ahead. This is the transonic case, $M=1$, the "sound barrier."

$\sin \theta =\frac{1}{M}=\frac{1}{2.5}=0.4\Rightarrow \theta ={\sin }^{-1}(0.4)=23.6^{\circ }.$ WHY ${\sin }^{-1}$: we know the ratio $\sin \theta$ and want the angle — ${\sin }^{-1}$ ("which angle has this sine?") undoes $\sin$. A faster rocket ($M=2.5>2$) gives a thinner cone than the $30^{\circ }$ of $M=2$.

Step 1 — get $M$. Invert $\sin \theta =1/M$: $M=\frac{1}{\sin 25^{\circ }}=\frac{1}{0.4226}=\mathrm{2.366.}$ Step 2 — get speed. By definition $M=v_{\text{obj}}/v_s$, so multiply back: $v_{\text{obj}}=M{v}_s=2.366\times 340\approx 804\text{m/s}.$ WHY two steps: the angle only encodes the ratio $M$; to recover an actual speed you must reintroduce $v_s$, the physical scale that $M$ threw away.

The cone angle depends only on $M$, so $M=1/\sin 25^{\circ }=2.366$ again. But $v_{\text{obj}}=M{v}_s=2.366\times 295\approx 698\text{m/s}.$ Lesson: identical cone shape ≠ identical speed. Cooler air → slower sound → the same $M$ means a lower actual speed.

Step 1 — Mach angle. $\sin \theta =1/1.8=0.5556\Rightarrow \theta =33.75^{\circ }$. Step 2 — where the cone wall hits the ground. The cone surface makes angle $\theta$ with the flight path. Looking at the right triangle (height $h$ opposite $\theta$, horizontal lag $x$ adjacent), the wall meets the ground a horizontal distance $x=\frac{h}{\tan \theta }=\frac{3000}{\tan 33.75^{\circ }}=\frac{3000}{0.6682}\approx 4490\text{m}$ behind the jet. WHY $\tan$: we relate the opposite side ($h$) to the adjacent side ($x$) of the same angle, and opposite-over-adjacent is $\tan$ — the natural tool when both legs matter and the hypotenuse doesn't. Step 3 — time. The jet's ground speed is $v_{\text{obj}}=M{v}_s=1.8\times 340=612\text{m/s}$. The cone trails the jet by $x$; the jet must fly that far before its wall reaches you: $t=\frac{x}{v_{\text{obj}}}=\frac{4490}{612}\approx 7.3\text{s}.$

As $M\to 1^{+}$, $\sin \theta =1/M\to 1$, so $\theta \to 90^{\circ }$ and $\tan \theta \to \infty$. Then $t=\frac{h}{v_{s}M\tan \theta }\to \frac{h}{v_s\cdot 1\cdot \infty }=0.$ Wait — check direction: the horizontal lag $x=h/\tan \theta \to 0$, so the cone wall is essentially vertical, right under the jet. The boom arrives almost the instant the jet is overhead. Physical meaning: at exactly $M=1$ the "cone" is a flat wall perpendicular to the ground track — there is no trailing lag. (The blow-up you might expect happens instead as $M\to \infty$: $\theta \to 0$, $\tan \theta \to 0$, $x\to \infty$ — the wall trails infinitely far behind.)

Warm layer: $M_1=700/340=2.059$, so $\sin {\theta }_1=1/2.059=0.4857$, ${\theta }_1=29.06^{\circ }$. Cold layer: $M_2=700/295=2.373$, so $\sin {\theta }_2=1/2.373=0.4214$, ${\theta }_2=24.93^{\circ }$. Reasoning: climbing into colder air lowers $v_s$, so at fixed true speed $M$ rises, which lowers $\sin \theta$, so the cone gets narrower ($29.06^{\circ }\to 24.93^{\circ }$). The rocket didn't speed up at all — the medium changed. This is exactly why engineers track $M$, not raw speed, when predicting wave drag and heating.

Same cone angle ⇒ same $M$: $M_Q=M_P=3.0$ (angle depends only on $M$). Let P's sound speed be $v_s$; then $v_{s,Q}=1.10v_s$. $\frac{v_Q}{v_P}=\frac{M_Q{v}_{s,Q}}{M_P{v}_s}=\frac{3.0\cdot 1.10v_s}{3.0\cdot v_s}=\mathrm{1.10.}$ So Q must fly $10\%$ faster than P to trace the identical cone. Insight: matching the shape forces matching $M$; matching $M$ in faster-sound air forces a proportionally higher true speed.

(a) A proper trailing cone needs $M>1$, i.e. $\sin \theta =1/M<1$. At $t=0$, $M=1.2>1$ already, so the cone exists as a trailing cone from the very start ($\theta <90^{\circ }$ for all $t\ge 0$). (b) At $t=0$: $M=1.2$, $\sin \theta =1/1.2=0.8333$, $\theta =56.44^{\circ }$. At $t=4$: $M=1.2+0.5(4)=3.2$, $\sin \theta =1/3.2=0.3125$, $\theta =18.21^{\circ }$. (c) As $t\to \infty$, $M\to \infty$, so $\sin \theta \to 0$ and $\theta \to 0^{\circ }$. Since $M$ increases monotonically, $\theta$ shrinks monotonically — the cone continuously narrows from $56.4^{\circ }$ toward a needle. It is widest at $t=0$ (slowest $M$) and never re-widens.

Smaller $\theta$ corresponds to larger $M$, so the limiting case is $\theta ={\theta }_{\min }=15^{\circ }$: $\sin 15^{\circ }=\frac{1}{M_{\max }}\Rightarrow M_{\max }=\frac{1}{\sin 15^{\circ }}=\frac{1}{0.2588}=\mathrm{3.86.}$ Interpretation: the capsule must stay below $M\approx 3.86$ to keep $\theta >15^{\circ }$. Above that speed the Mach cone lies down too flat, exactly the heating regime the shield can't survive. (Real re-entry is far more extreme, hypersonic — this is a simplified design gate.)

(a) $M=\frac{1}{\sin 28^{\circ }}=\frac{1}{0.4695}=\mathrm{2.130.}$ (b) $v_{\text{obj}}=M{v}_s=2.130\times 330=703\text{m/s}.$ (c) $x=\frac{h}{\tan \theta }=\frac{1500}{\tan 28^{\circ }}=\frac{1500}{0.5317}=2821\text{m}.$ (d) $t=\frac{x}{v_{\text{obj}}}=\frac{2821}{703}=4.01\text{s}.$ Why this order: angle → $M$ (invert $\sin$), $M$ → speed (multiply by $v_s$), speed + geometry → lag ($\tan$), lag ÷ speed → time. Each arrow uses exactly one earned tool.