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

3.4.19 · Physics › Rocket Flight Mechanics

$\beta$ HAI KYA?

YEH EXACT GROUPING KYUN AATA HAI? Kyunki motion ki equation mein, mass aur drag terms hamesha is ratio mein milke aate hain — kabhi alag nahi. Chalo isko derive karte hain.

KAISE: reentry deceleration ko first principles se derive karo

Ek body ko atmosphere mein straight-line path ke saath enter karte socho (yeh ek steep ballistic entry ke liye common pehla approximation hai). Flight direction ke saath Newton ka second law, sirf drag rakhte hue (peak-deceleration argument ke liye gravity component ignore karo):

$m\frac{dv}{dt}=-D=-\frac{1}{2}\rho v^2{C}_{D}A$

Yeh step kyun? Drag force hai $D=\frac{1}{2}\rho v^2{C}_{D}A$ — yeh air density $\rho$ aur $v^2$ ke saath badhti hai. Minus sign: drag motion ke khilaf hoti hai.

Dono sides ko $m$ se divide karo:

$\frac{dv}{dt}=-\frac{\rho v^2}{2}\cdot \frac{C_{D}A}{m}=-\frac{\rho v^2}{2\beta }$

Yeh step kyun? $m$ se divide karne par $\beta =m/(C_{D}A)$ naturally bahar aata hai. Yahi proof hai ki sirf combination $\beta$ matter karta hai, na ki $m$, $C_D$, $A$ alag-alag.

Isothermal atmosphere ke zariye altitude laao

Density exponentially girta hai: $\rho ={\rho }_0{e}^{-h/H}$, jahan $H$ scale height hai (~7–8 km Earth ke liye). Entry angle $\gamma$ (flight path horizontal se neeche) ke saath chain rule use karo, toh $dh/dt=-v\sin \gamma$:

$\frac{dv}{dh}=\frac{dv/dt}{dh/dt}=\frac{-\rho v^2/(2\beta )}{-v\sin \gamma }=\frac{\rho v}{2\beta \sin \gamma }$

Rearrange karke integrate karo ($v$ vs $\rho$):

$\frac{dv}{v}=\frac{\rho dh}{2\beta \sin \gamma }\Rightarrow \ln \frac{v}{v_e}=\frac{1}{2\beta \sin \gamma }{\int }_{\infty }^h\rho dh$

${\int }_h^{\infty }\rho dh=\rho H$ ke saath (exponential ka integral):

Peak deceleration (blunt = survivable kyun)

Deceleration magnitude $a=\frac{\rho v^2}{2\beta }$. $v(h)$ substitute karo aur $\rho$ ke upar maximize karo ($\frac{da}{d\rho }=0$ lo):

$a(\rho )=\frac{\rho v_e^2}{2\beta }\exp \left(-\frac{\rho H}{\beta \sin \gamma }\right)$

Derivative zero set karne par peak milta hai ${\rho }^{\ast }=\frac{\beta \sin \gamma }{H}$ par, aur:

WORKED EXAMPLES

COMMON MISTAKES

Active Recall

Connections

• Drag force and drag coefficient
• Exponential atmosphere and scale height H
• Allen–Eggers approximation
• Aerodynamic heating and thermal protection systems
• Terminal velocity (same $\rho v^2$ vs $\beta$ balance, lekin equilibrium par)
• Skip vs ballistic vs lifting reentry
• Newton's second law

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