Reentry mechanics — ballistic coefficient β = m - (C_D A): Worked examples

This is the drill page for the parent topic (index 3.4.19). The parent built the physics; here we exhaust the cases. Before any formula appears again, one reminder in plain words:

The scenario matrix

Every reentry problem is one cell of this table. If we work one example per cell, you have seen everything.

#	Case class	What is extreme here	Example that hits it
A	Big $β$ (cannonball)	inertia dominates → barely slows	Ex 1
B	Small $β$ (feather)	drag dominates → nearly stopped	Ex 2
C	Steep $γ \to 90°$	$sin γ \to 1$ , hardest brake	Ex 3
D	Shallow $γ \to 0°$	$sin γ \to 0$ , gentle	Ex 4
E	Degenerate $γ = 0°$	horizontal — formula blows up	Ex 5
F	$ρ \to 0$ (top of air)	limiting: $v \to v_{e}$	Ex 6
G	Peak-g, $β$ -independence proof	two bodies, same $a_{m a x}$	Ex 7
H	Altitude of peak (where $β$ does bite)	solve for $h^{*}$	Ex 8
I	Real-world word problem	Mars entry, different $H, ρ_{0}$	Ex 9
J	Exam twist	given $a_{m a x}$ , back out $γ$	Ex 10

Figure — Reentry mechanics — ballistic coefficient β = m - (C_D A)

The figure above is the map: the red curve is a big- $β$ cannonball (stays fast, brakes low), the blue curve a small- $β$ feather (brakes high). Every example below is a point on one of these curves — or a limiting edge of the plot.

Worked examples

Worked example Ex 1 — Cell A: the cannonball keeps its speed

A dense warhead has $β = 1 0^{5} kg/m^{2}$ . It hits sea-level air $ρ = 1.0 kg/m^{3}$ after a vertical dive ( $γ = 90°$ ), entering at $v_{e} = 7000 m/s$ . Use $H = 7000 m$ . What fraction of speed survives to sea level?

Forecast: guess — does a cannonball lose 5%, 50%, or 95% of its speed?

Compute the exponent $- \frac{ρ H}{2 β sin γ}$ . Why this step? The whole velocity profile lives in this one number; nothing else matters. $- \frac{1.0 \times 7000}{2 \times 1 0 ^{5} \times 1} = - \frac{7000}{2 \times 1 0 ^{5}} = - 0.035$
Exponentiate: $v = 7000 e^{- 0.035}$ . Why this step? $e^{- 0.035} \approx 0.9656$ — a tiny cut because the exponent is tiny. $v \approx 7000 \times 0.9656 = 6759 m/s$

Verify: It kept $\approx 96.6%$ of entry speed at the ground — "still screaming fast." Big $β$ ⇒ tiny exponent ⇒ almost no braking. Matches "Big Beta Bores Deep."

Worked example Ex 2 — Cell B: the feather is annihilated

Same conditions as Ex 1 ( $ρ = 1.0$ , $H = 7000$ , $γ = 90°$ , $v_{e} = 7000$ ) but a light blunt capsule with $β = 100 kg/m^{2}$ . Speed at sea level?

Forecast: 1000× smaller $β$ than the warhead — does the speed drop 1000× or far more?

Exponent: $- \frac{1.0 \times 7000}{2 \times 100 \times 1} = - 35$ . Why this step? Small $β$ multiplies the exponent hugely — this is the whole difference between the two bodies.
$v = 7000 e^{- 35}$ . Why this step? $e^{- 35} \approx 6.3 \times 1 0^{- 16}$ — astronomically small. $v \approx 7000 \times 6.3 \times 1 0^{- 16} \approx 4.4 \times 1 0^{- 12} m/s \approx 0$

Verify: Effectively stopped — "a feather in the pool." Note the exponent scaled linearly with $1/ β$ but the speed changed by 15 orders of magnitude: the exponential turns a 1000× input into a colossal output. That non-linearity is the whole point of Allen–Eggers approximation.

Worked example Ex 5 — Cell E: the degenerate

$γ = 0$ What does the theory say for a horizontal entry, $γ = 0°$ ?

Forecast: with no downward component, does the body ever descend? Does the formula give a number?

$sin 0° = 0$ sits in the denominator of the velocity exponent $- \frac{ρ H}{2 β sin γ}$ . Why this step? Division by zero — the formula is undefined, and that is physically correct: with $γ = 0$ , $d h / d t = - v sin γ = 0$ , so the body never changes altitude, never enters denser air.
$a_{m a x} = \frac{v _{e}^{2} \times 0}{2 eH} = 0$ . Why this step? No component of motion drives it into the atmosphere, so there is no build-up of $ρ$ and no peak.

Verify: Both formulas agree the case is degenerate: the velocity profile is undefined (never descends) and the "peak" deceleration is zero. Real bodies at $γ = 0$ skip off the atmosphere or orbit — see Skip vs ballistic vs lifting reentry. The math correctly refuses to give a finite ballistic answer.

Worked example Ex 8 — Cell H: the altitude where the peak occurs (where

$β$ finally matters) Peak deceleration happens at density $ρ^{*} = \frac{β sin γ}{H}$ . With $H = 7000$ , $γ = 30°$ , sea-level reference $ρ_{0} = 1.225 kg/m^{3}$ , find the peak altitude $h^{*} = H ln (ρ_{0} / ρ^{*})$ for warhead ( $β = 1 0^{5}$ ) vs capsule ( $β = 353$ ).

Forecast: which body's worst jolt happens lower (in denser, hotter air)?

Warhead $ρ^{*} = \frac{1 0 ^{5} \times 0.5}{7000} \approx 7.143 kg/m^{3}$ . Why this step? $ρ^{*}$ is bigger than sea-level $ρ_{0} = 1.225$ — meaning the peak would occur below sea level, i.e. the warhead never fully decelerates in the air. It hits the ground still braking. $h_{warhead}^{*} = 7000 ln (1.225/7.143) = 7000 \times (- 1.7636) \approx - 12345 m$ (negative ⇒ "peak" is theoretical, past the ground).
Capsule $ρ^{*} = \frac{353 \times 0.5}{7000} \approx 0.02521 kg/m^{3}$ . Why this step? Small $β$ → small $ρ^{*}$ → peak in thin, high, cool air. $h_{capsule}^{*} = 7000 ln (1.225/0.02521) = 7000 \times 3.882 \approx 27176 m$

Verify: Capsule peaks at $\approx 27 km$ altitude (cool, survivable heating); warhead's "peak" is below ground (it slams in still-fast). $β$ controls altitude, not g-magnitude — exactly the split promised in the parent and relevant to Aerodynamic heating and thermal protection systems.

Worked example Ex 9 — Cell I: real-world word problem, Mars entry

A Mars lander enters Mars' atmosphere at $v_{e} = 5900 m/s$ , $γ = 15°$ . Mars scale height $H_{Mars} = 11100 m$ (thinner-falling atmosphere). Peak deceleration in Earth-g?

Forecast: Mars air is famously thin — will the peak g be smaller than Earth's ~ $75 g$ cases?

Note the formula's inputs — $v_{e}, γ, H$ — no atmospheric density and no $β$ . Why this step? Even for Mars, only entry speed, angle, and scale height set the peak. Density $ρ_{0}$ cancels out of the maximization.
$a_{m a x} = \frac{( 5900 ) ^{2} sin 15°}{2 e ( 11100 )}$ , $sin 15° = 0.2588$ . $= \frac{3.481 \times 1 0 ^{7} \times 0.2588}{2 \times 2.71828 \times 11100} = \frac{9.009 \times 1 0 ^{6}}{60346} \approx 149.3 m/s^{2}$
In g: $149.3/9.81 \approx 15.2 g$ . Why this step? Bigger $H$ (denser air spread over more km) and small $sin 15°$ both soften the brake.

Verify: $\approx 15 g$ — far gentler than Earth's steep entries, because $H_{Mars}$ is larger and the angle is shallow. This is why Mars landers still need parachutes/retro-rockets: the atmosphere alone can't finish the job. Units check: $(m/s)^{2} / m = m/s^{2}$ ✓.

Worked example Ex 10 — Cell J: exam twist — back out the entry angle from a g-limit

Mission rule: a crewed capsule must never exceed $a_{m a x} = 4 g$ . Given $v_{e} = 7800 m/s$ , $H = 7000 m$ , what is the maximum allowed entry angle $γ$ ?

Forecast: to keep g tiny, must $γ$ be tiny too — a few degrees?

Rearrange $a_{m a x} = \frac{v _{e}^{2} sin γ}{2 eH}$ for $sin γ$ . Why this step? We know $a_{m a x}$ and want $γ$ — invert the formula. $sin γ = \frac{2 eH a _{m a x}}{v _{e}^{2}}$
$a_{m a x} = 4 \times 9.81 = 39.24 m/s^{2}$ . Plug in: $sin γ = \frac{2 \times 2.71828 \times 7000 \times 39.24}{( 7800 ) ^{2}} = \frac{1.4934 \times 1 0 ^{6}}{6.084 \times 1 0 ^{7}} = 0.02454$
$γ = arcsin (0.02454) \approx 1.406°$ . Why this step? $arcsin$ answers "which angle has this sine?" — see the $arctan$ -style reasoning in trig prerequisites.

Verify: $γ \approx 1.4°$ — an extremely shallow "grazing" entry, exactly how real crewed vehicles thread the reentry corridor. Sanity: at this $γ$ , $a_{m a x} = \frac{( 7800 ) ^{2} sin 1.406°}{2 e ( 7000 )} = \frac{6.084 \times 1 0 ^{7} \times 0.02454}{38056} \approx 39.2 m/s^{2} = 4 g$ ✓, round-trip consistent.

Recall

Recall Which cell asks "how deep before the worst jolt?" — and does

$β$ matter there? Cell H (altitude of peak). $β$ does matter here: $ρ^{*} = β sin γ / H$ , so higher $β$ ⇒ higher $ρ^{*}$ ⇒ lower, hotter peak altitude. It never changes the peak's magnitude (Cell G).

Recall What breaks the velocity formula, and why is that physically correct?

$γ = 0$ (Cell E): $sin 0 = 0$ in the denominator. Correct because a horizontal path never descends into denser air — no ballistic deceleration exists.

Cell A vs B in one line?

Big

β

keeps ~97% of speed to the ground; small

β

is stopped to essentially zero.

Does peak-g depend on

β

No — Ex 7 gives identical

75 g

for

β = 1 0^{5}

and

β = 353

Connections

Allen–Eggers approximation — the two formulas drilled here
Exponential atmosphere and scale height H — where $H$ and $ρ^{*}$ come from
Drag force and drag coefficient — the $\frac{1}{2} ρ v^{2} C_{D} A$ we integrated
Aerodynamic heating and thermal protection systems — why peak altitude (Ex 8) matters
Skip vs ballistic vs lifting reentry — the $γ = 0$ degenerate case (Ex 5)
Terminal velocity · Newton's second law