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

Before we compute anything, let us re-earn every symbol used below so nothing appears unexplained.

We only reuse tools already built upstairs:

Here $μ = G M_{\oplus} = 3.986 \times 1 0^{14} m^{3} / s^{2}$ , and Earth radius $R_{\oplus} = 6371$ km. Altitude $h$ and semi-major axis are linked by $a = R_{\oplus} + h$ for a circular orbit.

The scenario matrix

Every problem this topic can ask is one of these cells. The figure below shows the same eight cells laid out as a map — which input is extreme, and where each example lands. The examples then tag their cell.

Figure — Atmospheric drag — exponential atmosphere model, orbit decay

Cell	What changes	Physical meaning	Example
A. Small drag	$ρ$ tiny (high altitude)	negligible deceleration, slow decay	Ex 1
B. Large drag	$ρ$ big (low altitude)	violent deceleration near re-entry	Ex 2
C. Zero input	$ρ = 0$ (vacuum)	no drag → orbit is stable, degenerate case	Ex 3
D. Density ratio	compare two altitudes	exponential factor $e^{Δ h / H}$	Ex 4
E. Limiting / runaway	$ρ \to$ large as $a ↓$	$	da/dt
F. Word problem	real re-entry warning	translate news into numbers	Ex 6
G. Lifetime integral	integrate $d a / d t$	how long until it falls?	Ex 7
H. Exam twist	drag paradox in numbers	show $v$ rises after decay	Ex 8

Worked examples

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

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

Forecast: guess the order of magnitude — is it closer to $1 m/s^{2}$ or $1 0^{- 6} m/s^{2}$ ?

Use $a_{d r a g} = \frac{1}{2} B ρ v^{2}$ . Why this step? The magnitude of drag depends only on how draggy the body is ( $B$ ), how thick the air is ( $ρ$ ), and speed squared. No orbit info needed.
Compute $v^{2} = 761 3^{2} = 5.796 \times 1 0^{7} m^{2} / s^{2}$ . Why this step? Drag grows with the square of speed — this is where most of the force lives.
$a_{d r a g} = \frac{1}{2} (0.05) (6 \times 1 0^{- 13}) (5.796 \times 1 0^{7}) = 8.69 \times 1 0^{- 7} m/s^{2}$ . Why this step? Multiply it all out; the tiny $ρ$ crushes the answer down to a whisper.

Answer to the forecast: it is closer to $1 0^{- 6} m/s^{2}$ — about $8.69 \times 1 0^{- 7} m/s^{2}$ , nowhere near $1 m/s^{2}$ .

Verify: Units: $(m^{2} / kg) (kg/m^{3}) (m^{2} / s^{2}) = m/s^{2}$ . ✓ That's $\sim 9 \times 1 0^{- 8} g$ — invisible per second, but never rests.

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

Same CubeSat has decayed to $h = 150$ km, where $ρ = 2.1 \times 1 0^{- 9} kg/m^{3}$ and $v = 7815$ m/s. Same $B = 0.05 m^{2} / kg$ . Find the drag deceleration and compare to Ex 1.

Forecast: density jumped by a factor $\sim 3500$ . Does the deceleration jump by the same factor?

$v^{2} = 781 5^{2} = 6.107 \times 1 0^{7} m^{2} / s^{2}$ . Why this step? Speed only rose slightly (lower orbit is faster), so the $v^{2}$ term barely moves.
$a_{d r a g} = \frac{1}{2} (0.05) (2.1 \times 1 0^{- 9}) (6.107 \times 1 0^{7}) = 3.21 \times 1 0^{- 3} m/s^{2}$ . Why this step? Plug and chug.
Ratio to Ex 1: $3.21 \times 1 0^{- 3} /8.69 \times 1 0^{- 7} \approx 3690$ . Why this step? Almost entirely the density ratio — drag is a density-driven effect at fixed speed.

Answer to the forecast: Does deceleration jump by the same $\sim 3500 \times$ factor? Almost — it jumps by $\approx 3690 \times$ , slightly more than the density factor because $v^{2}$ also grew a little. So drag tracks density, plus a small speed bonus.

Verify: $ρ$ ratio $= 2.1 \times 1 0^{- 9} /6 \times 1 0^{- 13} = 3500$ ; our deceleration ratio $\approx 3690$ , the extra bit from the slightly higher $v^{2}$ . ✓ Consistent — drag explodes as you descend.

Worked example Ex 4 — Cell D: density ratio across two altitudes

In the thermosphere the scale height is $H = 60$ km. How much denser is the air at $h_{2} = 340$ km than at $h_{1} = 400$ km?

Forecast: dropping 60 km = exactly one scale height. So the ratio should be exactly $e^{?}$ .

Use $ρ (h) = ρ_{0} e^{- (h - h_{0}) / H}$ , form the ratio $\frac{ρ ( h _{2} )}{ρ ( h _{1} )} = e^{- (h_{2} - h_{1}) / H}$ . Why this step? The anchor pair $(ρ_{0}, h_{0})$ appears in both top and bottom, so taking a ratio cancels them — we never need to know the reference density or reference altitude.
$h_{2} - h_{1} = 340 - 400 = - 60$ km, so exponent $= - (- 60) /60 = + 1$ . Why this step? Going down by one scale height means $h_{2} < h_{1}$ , the exponent is positive, density grows.
Ratio $= e^{1} \approx 2.718$ . Why this step? That's the whole point of scale height — density multiplies by $e$ per $H$ of descent.

Verify: By the definition of scale height, one $H$ down = factor $e$ . $e^{1} = 2.718$ . ✓ If we had dropped $120$ km $= 2 H$ , ratio would be $e^{2} \approx 7.39$ .

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

Compare $∣ d a / d t ∣$ at $h = 400$ km ( $ρ_{1} = 5 \times 1 0^{- 12}$ , $a_{1} = 6771$ km) versus $h = 200$ km ( $ρ_{2} = 3 \times 1 0^{- 10}$ , $a_{2} = 6571$ km). $B = 0.02 m^{2} / kg$ . Show the runaway.

Forecast: $a$ barely changes (3% smaller), but $ρ$ jumps 60×. Which term controls $d a / d t$ ?

$\frac{d a}{d t}_{1} = B ρ_{1} μ a_{1} = (0.02) (5 \times 1 0^{- 12}) (3.986 \times 1 0^{14}) (6.771 \times 1 0^{6})$ . Why this step? Direct use of the decay rate; $μ a$ is the "orbital" part.
$(3.986 \times 1 0^{14}) (6.771 \times 1 0^{6}) = 2.699 \times 1 0^{21} = 5.195 \times 1 0^{10}$ , so rate $_{1} = (1 0^{- 13}) (5.195 \times 1 0^{10}) = 5.19 \times 1 0^{- 3}$ m/s $\approx 449$ m/day. Why this step? Convert to m/day (× 86400) for human intuition.
At $200$ km: $(3.986 \times 1 0^{14}) (6.571 \times 1 0^{6}) = 2.619 \times 1 0^{21} = 5.118 \times 1 0^{10}$ ; 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. Why this step? The $μ a$ term barely moved; the density did all the damage.

Verify: Rate ratio $= 0.307/5.19 \times 1 0^{- 3} \approx 59$ , essentially the density ratio $60$ . ✓ The decay rate is controlled almost entirely by $ρ$ — the runaway is a density effect.

The figure below shows this runaway visually: read it right-to-left as the satellite falls, and watch the black decay curve turn into the red cliff — a gentle glide becomes a plunge.

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

News: "A defunct 900 kg satellite, cross-section $4 m^{2}$ , drag coefficient $C_{D} = 2.2$ , is orbiting at $300$ km where $ρ = 2 \times 1 0^{- 11} kg/m^{3}$ ." Find its ballistic coefficient and current $d a / d t$ . $a = 6671$ km.

Forecast: heavy but broad-sided — is $B$ larger or smaller than a light CubeSat's $0.05$ ?

$B = \frac{C _{D} A}{m} = \frac{( 2.2 ) ( 4 )}{900} = 9.78 \times 1 0^{- 3} m^{2} / kg$ . Here $A = 4 m^{2}$ is the cross-sectional area — the shadow the satellite casts into the oncoming wind — and $m = 900$ kg its mass. Why this step? $B$ collapses shape ( $C_{D}$ ), size ( $A$ ) and mass ( $m$ ) into one "drag per kilogram" number. See Reentry Aerodynamics & Ballistic Coefficient.
$μ a = (3.986 \times 1 0^{14}) (6.671 \times 1 0^{6}) = 2.659 \times 1 0^{21} = 5.156 \times 1 0^{10}$ . Why this step? The orbital factor of the decay rate.
$\frac{d a}{d t} = - B ρ μ a = - (9.78 \times 1 0^{- 3}) (2 \times 1 0^{- 11}) (5.156 \times 1 0^{10}) = - 1.01 \times 1 0^{- 2}$ m/s $\approx - 871$ m/day. Why this step? Roughly a kilometre a day — this is why old satellites re-enter within months, not years.

Verify: $B = 9.78 \times 1 0^{- 3}$ is smaller than the CubeSat's $0.05$ — the heavy mass wins over the big area. Units of $d a / d t$ : $(m^{2} / kg) (kg/m^{3}) (m^{3} / s^{2}) (m) = (1/ m) \cdot (m^{2} / s) = m/s$ . ✓

Worked example Ex 7 — Cell G: lifetime integral

Estimate how long a satellite takes to decay from $a_{1} = 6771$ km to essentially re-entry, if we pretend density stays fixed at $ρ = 5 \times 1 0^{- 12}$ (a lower bound on speed). $B = 0.02 m^{2} / kg$ . Drop is $Δ a = 200$ km.

Forecast: at $\sim 450$ m/day (Ex 5), roughly how many days to shed 200 km?

Treat $d a / d t \approx - 449$ m/day as constant (crude, deliberately). Why this step? This is a lower bound on decay speed — real density rises, so real decay is faster. It brackets the answer.
Time $= \frac{Δ a}{∣ d a / d t ∣} = \frac{200 , 000 m}{449 m/day} \approx 445$ days. Why this step? Distance over rate = time, the simplest possible integral of a constant.

Verify: $445$ days $\approx 1.2$ years — a plausible upper limit. Because $ρ$ actually explodes (Ex 5 showed 26 km/day at 200 km), the true lifetime is far shorter, on the order of months. ✓ Constant- $ρ$ always overestimates lifetime — see the perturbation caveat.

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

A satellite decays from $a_{1} = 6771$ km to $a_{2} = 6671$ km. Using circular vis-viva $v = μ / a$ , show its speed increases even though drag removed energy.

Forecast: drag is friction — surely the satellite is slower afterward?

$v_{1} = μ / a_{1} = 3.986 \times 1 0^{14} /6.771 \times 1 0^{6} = 5.887 \times 1 0^{7} = 7673 m/s$ . Why this step? The starting circular speed.
$v_{2} = μ / a_{2} = 3.986 \times 1 0^{14} /6.671 \times 1 0^{6} = 5.975 \times 1 0^{7} = 7730 m/s$ . Why this step? The new, lower orbit's circular speed.
$Δ v = 7730 - 7673 = + 57 m/s$ — faster. Why this step? Smaller $a$ → larger $v$ ; the paradox lands in hard numbers.
Check energy: $ε_{1} = - μ /2 a_{1} = - 2.943 \times 1 0^{7}$ J/kg; $ε_{2} = - μ /2 a_{2} = - 2.987 \times 1 0^{7}$ J/kg. So $ε_{2} < ε_{1}$ — energy dropped by $4.4 \times 1 0^{5}$ J/kg. Why this step? Confirms energy decreased while speed rose.

Verify: Speed up ( $+ 57$ m/s) and energy down ( $Δ ε = - 4.4 \times 1 0^{5}$ J/kg) simultaneously — impossible on a table, natural in orbit, because falling closer to Earth trades potential energy into extra speed. ✓ This is the paradox from the parent note, now quantified.

Recall Which matrix cell is hardest to intuit, and why?

Cell C (zero density) and Cell H (paradox). Zero density shows drag physics smoothly returns to Keplerian orbits; the paradox shows losing energy raises speed. Both break everyday "friction slows things" instinct.

Recall Why does constant-density lifetime (Ex 7) overestimate the true lifetime?

Real density rises exponentially as altitude falls, so $∣ d a / d t ∣$ grows — the satellite spends its final orbits plunging far faster than the constant rate assumes.

Connections

Parent topic (Hinglish) — the formulas these examples exercise.
Vis-Viva Equation — the speed law behind Ex 8's paradox.
Orbital Energy & Specific Mechanical Energy — the energy bookkeeping in Ex 8.
Reentry Aerodynamics & Ballistic Coefficient — where $B$ comes from (Ex 6).
Hydrostatic Equilibrium (Atmospheres & Stars) — the origin of the exponential density (Ex 4).
Perturbations in Orbital Mechanics — drag as one perturbation among many (Ex 7).
Two-Body Problem — the drag-free limit recovered in Ex 3.