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

3.4.19 · D3 · Physics › Rocket Flight Mechanics › Reentry mechanics — ballistic coefficient β = m - (C_D A)

Yeh drill page hai parent topic (index 3.4.19) ke liye. Parent ne physics build ki; yahan hum saare cases exhaust karte hain. Koi bhi formula dobara aane se pehle, ek reminder plain words mein:

Scenario matrix

Har reentry problem is table ka ek cell hai. Agar hum ek example per cell karte hain, toh tumne sab kuch dekh liya.

#	Case class	Yahan kya extreme hai	Example jo ise hit karta hai
A	Big $β$ (cannonball)	inertia dominate karti hai → barely slows	Ex 1
B	Small $β$ (feather)	drag dominate karti hai → 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	do bodies, same $a_{m a x}$	Ex 7
H	Altitude of peak (jahan $β$ 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)

Upar wali figure map hai: red curve ek big- $β$ cannonball hai (fast rehta hai, neeche brake karta hai), blue curve ek small- $β$ feather (upar brake karta hai). Neeche har example in curves mein se kisi ek par ek point hai — ya plot ki koi limiting edge.

Worked examples

Worked example Ex 1 — Cell A: cannonball apni speed rakhta hai

Ek dense warhead ka $β = 1 0^{5} kg/m^{2}$ hai. Yeh vertical dive ( $γ = 90°$ ) ke baad sea-level air $ρ = 1.0 kg/m^{3}$ hit karta hai, $v_{e} = 7000 m/s$ se enter karta hai. $H = 7000 m$ use karo. Sea level tak speed ka kitna fraction survive karta hai?

Forecast: andaza lagao — kya ek cannonball apni speed ka 5%, 50%, ya 95% kho deta hai?

Exponent $- \frac{ρ H}{2 β sin γ}$ compute karo. Yeh step kyun? Poora velocity profile is ek number mein hi rehta hai; baaki kuch bhi matter nahi karta. $- \frac{1.0 \times 7000}{2 \times 1 0 ^{5} \times 1} = - \frac{7000}{2 \times 1 0 ^{5}} = - 0.035$
Exponentiate karo: $v = 7000 e^{- 0.035}$ . Yeh step kyun? $e^{- 0.035} \approx 0.9656$ — ek chota sa cut kyunki exponent chota hai. $v \approx 7000 \times 0.9656 = 6759 m/s$

Verify: Isne entry speed ka $\approx 96.6%$ ground par rakha — "abhi bhi screaming fast." Big $β$ ⇒ tiny exponent ⇒ almost no braking. "Big Beta Bores Deep" se match karta hai.

Worked example Ex 2 — Cell B: feather annihilate ho jaati hai

Ex 1 ke same conditions ( $ρ = 1.0$ , $H = 7000$ , $γ = 90°$ , $v_{e} = 7000$ ) lekin ek light blunt capsule ke saath $β = 100 kg/m^{2}$ ke saath. Sea level par speed?

Forecast: warhead se 1000× chota $β$ — kya speed 1000× girti hai ya bahut zyada?

Exponent: $- \frac{1.0 \times 7000}{2 \times 100 \times 1} = - 35$ . Yeh step kyun? Chota $β$ exponent ko bahut zyada multiply karta hai — yahi do bodies ke beech ka poora fark hai.
$v = 7000 e^{- 35}$ . Yeh step kyun? $e^{- 35} \approx 6.3 \times 1 0^{- 16}$ — 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 — "pool mein ek feather." Dhyan do ki exponent $1/ β$ ke saath linearly scale hua lekin speed 15 orders of magnitude se badli: exponential ek 1000× input ko ek colossal output mein badal deta hai. Yahi non-linearity Allen–Eggers approximation ka poora point hai.

Worked example Ex 5 — Cell E: degenerate

$γ = 0$ Theory kya kehti hai ek horizontal entry ke liye, $γ = 0°$ ?

Forecast: koi downward component nahi hone par, kya body kabhi descend karti hai? Kya formula ek number deta hai?

$sin 0° = 0$ velocity exponent $- \frac{ρ H}{2 β sin γ}$ ke denominator mein baithta hai. Yeh step kyun? Division by zero — formula undefined hai, aur yeh physically correct hai: $γ = 0$ ke saath, $d h / d t = - v sin γ = 0$ , isliye body kabhi altitude nahi badlati, kabhi denser air mein enter nahi karti.
$a_{m a x} = \frac{v _{e}^{2} \times 0}{2 eH} = 0$ . Yeh step kyun? Motion ka koi component use nahi atmosphere mein drive karta, isliye $ρ$ ka koi build-up nahi aur koi peak nahi.

Verify: Dono formulas agree karte hain ki case degenerate hai: velocity profile undefined hai (kabhi descend nahi karta) aur "peak" deceleration zero hai. Real bodies $γ = 0$ par atmosphere se skip karte hain ya orbit karte hain — dekho Skip vs ballistic vs lifting reentry. Math sahi se ek finite ballistic answer dene se mana kar deta hai.

Worked example Ex 8 — Cell H: jis altitude par peak hoti hai (jahan

$β$ finally matter karta hai) Peak deceleration density $ρ^{*} = \frac{β sin γ}{H}$ par hoti hai. $H = 7000$ , $γ = 30°$ , sea-level reference $ρ_{0} = 1.225 kg/m^{3}$ ke saath, warhead ( $β = 1 0^{5}$ ) vs capsule ( $β = 353$ ) ke liye peak altitude $h^{*} = H ln (ρ_{0} / ρ^{*})$ nikalte hain.

Forecast: kis body ka worst jolt neeche (denser, hotter air mein) hota hai?

Warhead $ρ^{*} = \frac{1 0 ^{5} \times 0.5}{7000} \approx 7.143 kg/m^{3}$ . Yeh step kyun? $ρ^{*}$ sea-level $ρ_{0} = 1.225$ se bada hai — matlab peak sea level ke neeche hogi, yaani warhead air mein kabhi fully decelerate nahi hota. Woh ground se takraata hai abhi bhi braking kar raha hota hai. $h_{warhead}^{*} = 7000 ln (1.225/7.143) = 7000 \times (- 1.7636) \approx - 12345 m$ (negative ⇒ "peak" theoretical hai, ground ke past).
Capsule $ρ^{*} = \frac{353 \times 0.5}{7000} \approx 0.02521 kg/m^{3}$ . Yeh step kyun? Chota $β$ → chota $ρ^{*}$ → peak thin, high, cool air mein. $h_{capsule}^{*} = 7000 ln (1.225/0.02521) = 7000 \times 3.882 \approx 27176 m$

Verify: Capsule $\approx 27 km$ altitude par peak karti hai (cool, survivable heating); warhead ki "peak" ground ke neeche hai (woh abhi-bhi-fast slam karta hai). $β$ altitude control karta hai, g-magnitude nahi — exactly woh split jo parent mein promise ki gayi thi aur Aerodynamic heating and thermal protection systems ke liye relevant hai.

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

Ek Mars lander Mars' atmosphere mein $v_{e} = 5900 m/s$ , $γ = 15°$ par enter karta hai. Mars scale height $H_{Mars} = 11100 m$ (thinner-falling atmosphere). Peak deceleration in Earth-g?

Forecast: Mars ki air famously thin hai — kya peak g Earth ke ~ $75 g$ cases se smaller hoga?

Formula ke inputs note karo — $v_{e}, γ, H$ — koi atmospheric density nahi aur koi $β$ nahi. Yeh step kyun? Mars ke liye bhi, sirf entry speed, angle, aur scale height hi peak set karte hain. Density $ρ_{0}$ maximization se cancel ho jaati hai.
$a_{m a x} = \frac{( 5900 ) ^{2} sin 15°}{2 e ( 11100 )}$ , $sin 15° = 0.2588$ . $= \frac{3.481 \times 1 0 ^{7} \times 0.2588}{2 \times 2.71828 \times 11100} = \frac{9.009 \times 1 0 ^{6}}{60346} \approx 149.3 m/s^{2}$
g mein: $149.3/9.81 \approx 15.2 g$ . Yeh step kyun? Bada $H$ (denser air zyada km mein spread) aur chota $sin 15°$ dono brake soften karte hain.

Verify: $\approx 15 g$ — Earth ki steep entries se far gentler, kyunki $H_{Mars}$ larger hai aur angle shallow hai. Yahi wajah hai ki Mars landers ko abhi bhi parachutes/retro-rockets chahiye: atmosphere akela kaam nahi kar sakta. Units check: $(m/s)^{2} / m = m/s^{2}$ ✓.

Worked example Ex 10 — Cell J: exam twist — g-limit se entry angle back out karo

Mission rule: ek crewed capsule kabhi $a_{m a x} = 4 g$ exceed nahi kar sakti. Given $v_{e} = 7800 m/s$ , $H = 7000 m$ , maximum allowed entry angle $γ$ kya hai?

Forecast: g tiny rakhne ke liye, kya $γ$ bhi tiny hona chahiye — kuch degrees?

$a_{m a x} = \frac{v _{e}^{2} sin γ}{2 eH}$ ko $sin γ$ ke liye rearrange karo. Yeh step kyun? Hume $a_{m a x}$ pata hai aur $γ$ chahiye — formula invert karo. $sin γ = \frac{2 eH a _{m a x}}{v _{e}^{2}}$
$a_{m a x} = 4 \times 9.81 = 39.24 m/s^{2}$ . Plug in karo: $sin γ = \frac{2 \times 2.71828 \times 7000 \times 39.24}{( 7800 ) ^{2}} = \frac{1.4934 \times 1 0 ^{6}}{6.084 \times 1 0 ^{7}} = 0.02454$
$γ = arcsin (0.02454) \approx 1.406°$ . Yeh step kyun? $arcsin$ ka jawab hai "kis angle ki yeh sine hai?" — trig prerequisites mein $arctan$ -style reasoning dekho.

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

Recall

Recall Kaun sa cell poochta hai "worst jolt se pehle kitna deep?" — aur kya

$β$ wahan matter karta hai? Cell H (altitude of peak). $β$ yahan does matter karta hai: $ρ^{*} = β sin γ / H$ , isliye zyada $β$ ⇒ zyada $ρ^{*}$ ⇒ lower, hotter peak altitude. Yeh peak ki magnitude kabhi nahi badlata (Cell G).

Recall Velocity formula kya break karta hai, aur yeh physically correct kyun hai?

$γ = 0$ (Cell E): denominator mein $sin 0 = 0$ . Correct hai kyunki ek horizontal path kabhi denser air mein descend nahi karta — koi ballistic deceleration exist nahi karti.

Cell A vs B ek line mein?

Big

β

ground tak ~97% speed rakhta hai; small

β

essentially zero par rok diya jaata hai.

Kya peak-g

β

par depend karta hai?

Nahi — Ex 7

β = 1 0^{5}

aur

β = 353

dono ke liye identical

75 g

deta hai.

Connections

Allen–Eggers approximation — yahan drill kiye gaye do formulas
Exponential atmosphere and scale height H — jahan se $H$ aur $ρ^{*}$ aate hain
Drag force and drag coefficient — woh $\frac{1}{2} ρ v^{2} C_{D} A$ jo humne integrate kiya
Aerodynamic heating and thermal protection systems — kyun peak altitude (Ex 8) matter karta hai
Skip vs ballistic vs lifting reentry — $γ = 0$ degenerate case (Ex 5)
Terminal velocity · Newton's second law