Atmospheric drag — exponential atmosphere model, orbit decay: Worked examples

3.2.34 · D3 · Physics › Orbital Mechanics & Astrodynamics › Atmospheric drag — exponential atmosphere model, orbit decay

Kuch bhi compute karne se pehle, neeche use hone wale har symbol ko dobara samjhte hain taaki kuch bhi unexplained na rahe.

Definition Pehle yeh chaar symbols chahiye

Ballistic coefficient $B = \frac{C _{D} A}{m}$ : ek single "drag per kilogram" number.
- $C_{D}$ = drag coefficient (shape kitni bluntly hawa pakadti hai, dimensionless, $\approx 2.2$ ek tumbling satellite ke liye).
- $A$ = woh cross-sectional area jo body aane wali hawa ke saamne present karti hai (m², jaise hawa mein uski shadow).
- $m$ = mass (kg).
- Bada $B$ (halka aur choda) ⇒ jaldi decay hota hai; chhota $B$ (bhaari aur patla) ⇒ dheere decay hota hai.
Reference density $ρ_{0}$ : ek chosen altitude $h_{0}$ par maapi gayi air density (woh "anchor point" jo hum jaante hain).
Reference altitude $h_{0}$ : woh height jahan $ρ_{0}$ measure ki gayi — exponential curve is point par pin hoti hai.
Scale height $H$ : wo vertical drop jitne mein density factor $e \approx 2.718$ se kam ho jaati hai.

Hum sirf wahi tools use karte hain jo upar banaye ja chuke hain:

Yahan $μ = G M_{\oplus} = 3.986 \times 1 0^{14} m^{3} / s^{2}$ , aur Earth radius $R_{\oplus} = 6371$ km. Altitude $h$ aur semi-major axis $a = R_{\oplus} + h$ se jude hain circular orbit ke liye.

Scenario matrix

Is topic ke har possible problem ka ek cell hai. Neeche ki figure inhi aath cells ko map ki tarah dikhati hai — kaunsa input extreme hai, aur har example kahan land karta hai. Examples phir apna cell tag karte hain.

Figure — Atmospheric drag — exponential atmosphere model, orbit decay

Cell	Kya badlta hai	Physical meaning	Example
A. Small drag	$ρ$ tiny (high altitude)	negligible deceleration, slow decay	Ex 1
B. Large drag	$ρ$ bada (low altitude)	violent deceleration near re-entry	Ex 2
C. Zero input	$ρ = 0$ (vacuum)	koi drag nahi → orbit stable hai, degenerate case	Ex 3
D. Density ratio	do altitudes compare karo	exponential factor $e^{Δ h / H}$	Ex 4
E. Limiting / runaway	$ρ \to$ bada jab $a ↓$	$	da/dt
F. Word problem	real re-entry warning	news ko numbers mein translate karo	Ex 6
G. Lifetime integral	$d a / d t$ integrate karo	kitne time mein giregi?	Ex 7
H. Exam twist	drag paradox in numbers	dikhaao ki decay ke baad $v$ badhta hai	Ex 8

Worked examples

Worked example Ex 1 — Cell A: small drag, high orbit

Ek CubeSat $h = 500$ km par, ballistic coefficient $B = 0.05 m^{2} / kg$ (yaad karo $B = C_{D} A / m$ , drag area per kg), local density $ρ = 6 \times 1 0^{- 13} kg/m^{3}$ , orbital speed $v = 7613$ m/s. Drag deceleration nikalo.

Forecast: order of magnitude guess karo — yeh $1 m/s^{2}$ ke kareeb hogi ya $1 0^{- 6} m/s^{2}$ ke?

$a_{d r a g} = \frac{1}{2} B ρ v^{2}$ use karo. Yeh step kyun? Drag ki magnitude sirf is par depend karti hai ki body kitni draggy hai ( $B$ ), hawa kitni thick hai ( $ρ$ ), aur speed squared. Koi orbit info nahi chahiye.
$v^{2} = 761 3^{2} = 5.796 \times 1 0^{7} m^{2} / s^{2}$ compute karo. Yeh step kyun? Drag speed ke square ke saath badhta hai — yahan hi zyaadatar force rehti hai.
$a_{d r a g} = \frac{1}{2} (0.05) (6 \times 1 0^{- 13}) (5.796 \times 1 0^{7}) = 8.69 \times 1 0^{- 7} m/s^{2}$ . Yeh step kyun? Sab multiply karo; tiny $ρ$ answer ko whisper tak daba deta hai.

Forecast ka jawab: yeh $1 0^{- 6} m/s^{2}$ ke kareeb hai — lagbhag $8.69 \times 1 0^{- 7} m/s^{2}$ , $1 m/s^{2}$ ke aaspaas bilkul nahi.

Verify: Units: $(m^{2} / kg) (kg/m^{3}) (m^{2} / s^{2}) = m/s^{2}$ . ✓ Yeh $\sim 9 \times 1 0^{- 8} g$ hai — per second invisible, lekin kabhi rest nahi leta.

Worked example Ex 2 — Cell B: large drag, low orbit

Wahi CubeSat $h = 150$ km tak decay ho gaya, jahan $ρ = 2.1 \times 1 0^{- 9} kg/m^{3}$ aur $v = 7815$ m/s hai. Same $B = 0.05 m^{2} / kg$ . Drag deceleration nikalo aur Ex 1 se compare karo.

Forecast: density $\sim 3500$ factor se jump ki. Kya deceleration bhi usi factor se jump karegi?

$v^{2} = 781 5^{2} = 6.107 \times 1 0^{7} m^{2} / s^{2}$ . Yeh step kyun? Speed thodi hi badhi (lower orbit thoda faster hoti hai), isliye $v^{2}$ term barely hilti hai.
$a_{d r a g} = \frac{1}{2} (0.05) (2.1 \times 1 0^{- 9}) (6.107 \times 1 0^{7}) = 3.21 \times 1 0^{- 3} m/s^{2}$ . Yeh step kyun? Plug and chug.
Ex 1 se ratio: $3.21 \times 1 0^{- 3} /8.69 \times 1 0^{- 7} \approx 3690$ . Yeh step kyun? Almost poora density ratio hai — drag fixed speed par ek density-driven effect hai.

Forecast ka jawab: Kya deceleration same $\sim 3500 \times$ factor se jump karti hai? Almost — yeh $\approx 3690 \times$ jump karti hai, density factor se thoda zyaada kyunki $v^{2}$ bhi thoda bada hua. Toh drag density track karta hai, plus ek small speed bonus.

Verify: $ρ$ ratio $= 2.1 \times 1 0^{- 9} /6 \times 1 0^{- 13} = 3500$ ; hamara deceleration ratio $\approx 3690$ , extra bit thodi zyaada $v^{2}$ se. ✓ Consistent — jaise-jaise neeche aate ho drag explode karta hai.

Worked example Ex 4 — Cell D: do altitudes mein density ratio

Thermosphere mein scale height $H = 60$ km hai. $h_{2} = 340$ km par hawa $h_{1} = 400$ km se kitni dense hai?

Forecast: 60 km neeche jaana = exactly ek scale height. Toh ratio exactly $e^{?}$ hona chahiye.

$ρ (h) = ρ_{0} e^{- (h - h_{0}) / H}$ use karo, ratio banao $\frac{ρ ( h _{2} )}{ρ ( h _{1} )} = e^{- (h_{2} - h_{1}) / H}$ . Yeh step kyun? Anchor pair $(ρ_{0}, h_{0})$ upar aur neeche dono mein appear hota hai, isliye ratio lene par cancel ho jaate hain — reference density ya reference altitude jaanna zaroori nahi.
$h_{2} - h_{1} = 340 - 400 = - 60$ km, toh exponent $= - (- 60) /60 = + 1$ . Yeh step kyun? Ek scale height neeche jaana matlab $h_{2} < h_{1}$ , exponent positive hai, density badhti hai.
Ratio $= e^{1} \approx 2.718$ . Yeh step kyun? Scale height ka yahi toh point hai — $H$ ke ek descent par density $e$ se multiply hoti hai.

Verify: Scale height ki definition se, ek $H$ neeche = factor $e$ . $e^{1} = 2.718$ . ✓ Agar $120$ km $= 2 H$ neeche jaate, ratio $e^{2} \approx 7.39$ hota.

Worked example Ex 5 — Cell E: limiting / runaway decay

$∣ d a / d t ∣$ compare karo $h = 400$ km ( $ρ_{1} = 5 \times 1 0^{- 12}$ , $a_{1} = 6771$ km) par aur $h = 200$ km ( $ρ_{2} = 3 \times 1 0^{- 10}$ , $a_{2} = 6571$ km) par. $B = 0.02 m^{2} / kg$ . Runaway dikhao.

Forecast: $a$ barely badla (3% chhota), lekin $ρ$ 60× jump kiya. Kaunsa term $d a / d t$ control karta hai?

$\frac{d a}{d t}_{1} = B ρ_{1} μ a_{1} = (0.02) (5 \times 1 0^{- 12}) (3.986 \times 1 0^{14}) (6.771 \times 1 0^{6})$ . Yeh step kyun? Decay rate ka direct use; $μ a$ "orbital" part hai.
$(3.986 \times 1 0^{14}) (6.771 \times 1 0^{6}) = 2.699 \times 1 0^{21} = 5.195 \times 1 0^{10}$ , toh rate $_{1} = (1 0^{- 13}) (5.195 \times 1 0^{10}) = 5.19 \times 1 0^{- 3}$ m/s $\approx 449$ m/day. Yeh step kyun? Human intuition ke liye m/day mein convert karo (× 86400).
$200$ km par: $(3.986 \times 1 0^{14}) (6.571 \times 1 0^{6}) = 2.619 \times 1 0^{21} = 5.118 \times 1 0^{10}$ ; rate $_{2} = (0.02) (3 \times 1 0^{- 10}) (5.118 \times 1 0^{10}) = 0.307$ m/s $\approx 26, 500$ m/day $= 26.5$ km/day. Yeh step kyun? $μ a$ term barely hila; density ne saara damage kiya.

Verify: Rate ratio $= 0.307/5.19 \times 1 0^{- 3} \approx 59$ , essentially density ratio $60$ . ✓ Decay rate almost poora $ρ$ se control hota hai — runaway ek density effect hai.

Neeche ki figure yeh runaway visually dikhati hai: ise right-to-left padho jaise satellite girta hai, aur dekho kaise black decay curve red cliff ban jaati hai — ek gentle glide ek plunge ban jaata hai.

Worked example Ex 6 — Cell F: real-world word problem

News: "Ek defunct 900 kg satellite, cross-section $4 m^{2}$ , drag coefficient $C_{D} = 2.2$ , $300$ km par orbit kar raha hai jahan $ρ = 2 \times 1 0^{- 11} kg/m^{3}$ hai." Iska ballistic coefficient aur current $d a / d t$ nikalo. $a = 6671$ km.

Forecast: bhaari lekin broad-sided — kya $B$ ek light CubeSat ke $0.05$ se bada hoga ya chhota?

$B = \frac{C _{D} A}{m} = \frac{( 2.2 ) ( 4 )}{900} = 9.78 \times 1 0^{- 3} m^{2} / kg$ . Yahan $A = 4 m^{2}$ cross-sectional area hai — satellite ki aane wali hawa mein shadow — aur $m = 900$ kg iska mass. Yeh step kyun? $B$ shape ( $C_{D}$ ), size ( $A$ ) aur mass ( $m$ ) ko ek "drag per kilogram" number mein compress karta hai. Reentry Aerodynamics & Ballistic Coefficient dekho.
$μ a = (3.986 \times 1 0^{14}) (6.671 \times 1 0^{6}) = 2.659 \times 1 0^{21} = 5.156 \times 1 0^{10}$ . Yeh step kyun? Decay rate ka orbital factor.
$\frac{d a}{d t} = - B ρ μ a = - (9.78 \times 1 0^{- 3}) (2 \times 1 0^{- 11}) (5.156 \times 1 0^{10}) = - 1.01 \times 1 0^{- 2}$ m/s $\approx - 871$ m/day. Yeh step kyun? Roughly ek kilometre per day — isliye purane satellites mahino mein re-enter karte hain, salon mein nahi.

Verify: $B = 9.78 \times 1 0^{- 3}$ CubeSat ke $0.05$ se chhota hai — bhaari mass bade area par haavi hoti hai. $d a / d t$ ke units: $(m^{2} / kg) (kg/m^{3}) (m^{3} / s^{2}) (m) = (1/ m) \cdot (m^{2} / s) = m/s$ . ✓

Worked example Ex 7 — Cell G: lifetime integral

Estimate karo ki ek satellite $a_{1} = 6771$ km se essentially re-entry tak kitne time mein decay hogi, agar hum density ko fixed maanein $ρ = 5 \times 1 0^{- 12}$ par (speed ki lower bound). $B = 0.02 m^{2} / kg$ . Drop $Δ a = 200$ km hai.

Forecast: $\sim 450$ m/day (Ex 5) par, 200 km shed karne mein roughly kitne din?

$d a / d t \approx - 449$ m/day ko constant maano (deliberately crude). Yeh step kyun? Yeh decay speed ki lower bound hai — real density badhti hai, isliye real decay faster hoti hai. Yeh answer ko bracket karta hai.
Time $= \frac{Δ a}{∣ d a / d t ∣} = \frac{200 , 000 m}{449 m/day} \approx 445$ days. Yeh step kyun? Distance over rate = time, constant ka sabse simple possible integral.

Verify: $445$ days $\approx 1.2$ saal — ek plausible upper limit. Kyunki $ρ$ actually explode hota hai (Ex 5 ne 26 km/day 200 km par dikhaya), true lifetime far shorter hai, months ke order mein. ✓ Constant- $ρ$ hamesha lifetime overestimate karta hai — perturbation caveat dekho.

Worked example Ex 8 — Cell H: exam twist (numbers mein drag paradox)

Ek satellite $a_{1} = 6771$ km se $a_{2} = 6671$ km tak decay hoti hai. Circular vis-viva $v = μ / a$ use karke dikhao ki iska speed badhta hai bhale hi drag ne energy churaai.

Forecast: drag toh friction hai — kya satellite baad mein slower nahi honi chahiye?

$v_{1} = μ / a_{1} = 3.986 \times 1 0^{14} /6.771 \times 1 0^{6} = 5.887 \times 1 0^{7} = 7673 m/s$ . Yeh step kyun? Starting circular speed.
$v_{2} = μ / a_{2} = 3.986 \times 1 0^{14} /6.671 \times 1 0^{6} = 5.975 \times 1 0^{7} = 7730 m/s$ . Yeh step kyun? Nayi, neeche wali orbit ki circular speed.
$Δ v = 7730 - 7673 = + 57 m/s$ — faster. Yeh step kyun? Chhota $a$ → bada $v$ ; paradox hard numbers mein land karta hai.
Energy check: $ε_{1} = - μ /2 a_{1} = - 2.943 \times 1 0^{7}$ J/kg; $ε_{2} = - μ /2 a_{2} = - 2.987 \times 1 0^{7}$ J/kg. Toh $ε_{2} < ε_{1}$ — energy $4.4 \times 1 0^{5}$ J/kg se giri. Yeh step kyun? Confirm karta hai ki energy giri jabki speed badhi.

Verify: Speed up ( $+ 57$ m/s) aur energy down ( $Δ ε = - 4.4 \times 1 0^{5}$ J/kg) simultaneously — roz ki zindagi mein impossible, orbit mein natural, kyunki Earth ke kareeb girne par potential energy extra speed mein convert hoti hai. ✓ Yeh parent note ka paradox hai, ab quantified.

Recall Matrix mein kaun sa cell intuition se sabse mushkil hai, aur kyun?

Cell C (zero density) aur Cell H (paradox). Zero density dikhata hai ki drag physics smoothly Keplerian orbits mein wapas aa jaati hai; paradox dikhata hai ki energy kho kar speed badhti hai. Dono "friction cheez dheeri karti hai" wali roz ki instinct tod dete hain.

Recall Constant-density lifetime (Ex 7) true lifetime kyun overestimate karta hai?

Real density exponentially badhti hai jaise altitude girती है, isliye $∣ d a / d t ∣$ badhta hai — satellite apne last orbits mein constant rate se kahi zyaada tezi se gir raha hota hai.

Connections

Parent topic (Hinglish) — woh formulas jo yeh examples exercise karte hain.
Vis-Viva Equation — Ex 8 ke paradox ke peeche speed law.
Orbital Energy & Specific Mechanical Energy — Ex 8 mein energy bookkeeping.
Reentry Aerodynamics & Ballistic Coefficient — $B$ kahan se aata hai (Ex 6).
Hydrostatic Equilibrium (Atmospheres & Stars) — exponential density ki origin (Ex 4).
Perturbations in Orbital Mechanics — drag as ek perturbation among many (Ex 7).
Two-Body Problem — Ex 3 mein recover hone wala drag-free limit.