Atmospheric drag — exponential atmosphere model, orbit decay

3.2.34 · Physics › Orbital Mechanics & Astrodynamics

Drag kyun matter karta hai aur WO HAI KYA?

KYA HAI: Drag ek force hai jo fluid mein motion ko oppose karti hai (yahan, residual upper atmosphere).

YE SUBTLE KYUN HAI: Orbit mein, kinetic aur potential energy linked hain. Energy hatane se orbit chhoti hoti hai, aur chhoti orbit ka matlab zyada speed hoti hai. Toh drag = decay = eventual reentry.

KAISE: drag force first principles se

Imagine karo satellite (cross-section $A$ ) density $ρ$ waali air mein speed $v$ se guzar rahi hai. Time $d t$ mein wo length $v d t$ ki ek tube sweep karti hai, jisme mass hota hai:

$d m_{ai r} = ρ A v d t$

Agar wo us gas ko (roughly) speed $v$ deti hai, toh momentum per time diya jaata hai:

$F \sim \frac{d m _{ai r}}{d t} v = ρ A v^{2}$

Real bodies sirf ek fraction transfer karti hain, jo drag coefficient $C_{D}$ se capture hota hai (aur standard aerodynamic definition se $\frac{1}{2}$ ka factor aata hai):

Ye step kyun? Humne $v v$ likha (nahi $v^{2}$ ) taaki direction $- \overset{v}{^}$ ho jabki magnitude $v^{2}$ se scale kare.

KAISE: exponential atmosphere model

Exponential kyun? Hydrostatic equilibrium mein air ka ek slab gravity ke against pressure difference se tika rehta hai:

$d P = - ρ g d h$

Upper air ko (roughly) constant temperature $T$ par ideal gas treat karte hue:

$P = \frac{ρ k _{B} T}{m _{m o l}} \Rightarrow d P = \frac{k _{B} T}{m _{m o l}} d ρ$

Substitute karo:

$\frac{k _{B} T}{m _{m o l}} d ρ = - ρ g d h \Rightarrow \frac{d ρ}{ρ} = - \frac{m _{m o l} g}{k _{B} T} d h$

Integrate karo → exponential:

Ye step kyun? Clean $d ρ / ρ = - d h / H$ directly exponential mein integrate ho jaata hai — density tezi se girta hai, isliye drag essentially perigee ke paas ek surface phenomenon hai.

KAISE: orbit decay — energy aur altitude loss

Near-circular orbit ke liye, total specific energy:

$ε = - \frac{μ}{2 a}, v = \frac{μ}{a} (circular)$

jahan $μ = G M_{\oplus}$ aur $a$ = semi-major axis.

Drag per unit time negative work karta hai (power):

$\frac{d ε}{d t} = a_{d r a g} \cdot v = - \frac{1}{2} B ρ v^{2} \cdot v = - \frac{1}{2} B ρ v^{3}$

Ye step kyun? Power = force · velocity; drag $v$ ke anti-parallel hai, toh work negative hai.

Ab $ε$ ko $a$ se link karo: kyunki $ε = - μ /2 a$ ,

$\frac{d ε}{d t} = \frac{μ}{2 a ^{2}} \frac{d a}{d t}$

Equal set karo aur $v^{2} = μ / a$ use karo, toh $v^{3} = (μ / a)^{3/2}$ :

$\frac{μ}{2 a ^{2}} \frac{d a}{d t} = - \frac{1}{2} B ρ (\frac{μ}{a})^{3/2}$

Figure — Atmospheric drag — exponential atmosphere model, orbit decay

Worked examples

Common mistakes

Active recall

Recall Flip karo mujhe: drag satellite ko faster kyun banata hai?

Drag energy remove karta hai → chhota $a$ . Circular speed $v = μ / a$ badhti hai jab $a$ girta hai. Gravity lost altitude ko extra speed mein convert karta hai.

Recall 12-saal ke bachche ko explain karo (Feynman)

Imagine karo tum ek bowl ke top rim ke paas skate kar rahe ho. Thodi si sticky air tum par rubbing karti hai aur teri zip chura leti hai. Lekin energy lose karna tumhe bowl mein neeche slide kara deta hai — aur wahan walls zyada steep hain, toh actually tum faster ghoomte ho! Jitna neeche jaoge, utni hi thick aur sticky air hogi, toh tum tezi se spiral karte neeche jaate ho jab tak — whoosh — tum bottom par jal jaate ho. Yahi ek satellite ka sky se girna hai.

Ballistic coefficient B kya hai?

B = C_{D} A / m

— drag area per unit mass; bada B jaldi decay karta hai.

Drag acceleration vector likho.

a_{d r a g} = - \frac{1}{2} C_{D} \frac{A}{m} ρ v v = - \frac{1}{2} B ρ v v

Atmospheric density altitude mein exponential kyun hai?

Hydrostatic equilibrium

d P = - ρ g d h

+ constant T par ideal gas deta hai

d ρ / ρ = - d h / H

, integrate karne par

ρ = ρ_{0} e^{- (h - h_{0}) / H}

Scale height define karo aur iska formula do.

Altitude jitne par density factor e se drop ho;

H = k_{B} T / (m_{m o l} g)

Near-circular orbit ke liye orbit-decay rate da/dt kya hai?

d a / d t = - B ρ μ a

(negative → shrinks).

Decaying satellite energy lose karne ke bawajood speed up kyun karta hai?

v = μ / a

; drag a lower karta hai, toh v badhta hai chahe total energy

ε = - μ /2 a

decrease ho.

Decay ant mein accelerate kyun karta hai?

Jab a shrinks hota hai, altitude drop hoti hai, ρ exponentially badhta hai, toh |da/dt|∝ρ blow up hota hai → runaway reentry.

a ke terms mein specific circular-orbit energy?

ε = - μ / (2 a)

Drag power (energy loss rate)?

d ε / d t = - \frac{1}{2} B ρ v^{3}

Connections

Two-Body Problem — $v = μ / a$ aur $ε = - μ /2 a$ provide karta hai jo yahan use hua.
Vis-Viva Equation — drag paradox ke peeche general speed–radius relation.
Orbital Energy & Specific Mechanical Energy — woh quantity jo drag remove karta hai.
Perturbations in Orbital Mechanics — drag as a non-conservative perturbing force.
Reentry Aerodynamics & Ballistic Coefficient — jahan decay khatam hoti hai.
Hydrostatic Equilibrium (Atmospheres & Stars) — exponential model ki origin.